Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5

Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities

Page 48 Essential Question Answer

An inequality with a closed dot on the number line and a boundary point represented by the “or equal” component of the symbols ≤≤ and≥≥. The sign (∞) denotes an unlimited interval to the right. Use the symbols< for “less than” and > for “greater than” to express ordering relationships.

We can use inequalities to describe intervals on the real number line as a closed dot on the number line and a square bracket in interval notation imply inclusive inequalities with the “or equal to” component.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Page 49 Exercise 2 Answer

The given graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 1

We represent this inequality using
x≤−6 or x>3

The required solution is x≤−6 or x>3 and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 2

 

The given graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 3

We represent this inequality using
x<−5 or x≥4

The required solution is x<−5 or x≥4 and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 4

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5

The given graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 5

We represent this inequality using
x≤−4 or x≥5

The required solution isx≤−4 or x≥5 and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 6

 

The given graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 7

We represent this inequality using
x<−3 or x>6

The required solution is x<−3 or x>6 and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 8

We use “or” in these inequalities to represent the union of the graph.

We use “or” in these inequalities to represent the union of the graph.

Big Ideas Math Algebra 1 Chapter 2 Exercise 2.5 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5 page 49 Exercise 3 Answer

We use combinations of inequalities to represent these graphs on the number line.
In addition, we use “and” to get the intersection of the two inequalities, and “or” to get the union of the two inequalities.

We use combinations of inequalities using “and” and “or” statements to represent these graphs on the number line.

 

Page 51 Exercise 2 Answer

Given: A number less than−2 and greater than or equal to 2.
To find Respective inequality.
Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 9

We represent this inequality as follows
​u<−2
u≥2
u<−2 or u≥2

The plot can be plotted as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 10

 

The obtained inequality is −2<u≤2 and the plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 11

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5 Page 51 Exercise 4 Answer

Given: A number is more than −4 and at most \(-6 \frac{1}{2}\).
To find Respective inequality.
Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 12

We represent this inequality using the following
c>−4 or c≤\(-6 \frac{1}{2}\)

The plot can be

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 13

The obtained inequality is c>−4 or c≤−\(-6 \frac{1}{2}\) and the plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 14

Solving Linear Inequalities Exercise 2.5 Big Ideas Math

Page 51 Exercise 5 Answer

Given: A number is no less than −1.5 and less than 5.3.
To find Respective inequality.
Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 15

We represent this graph using the following inequalities:
c≥−1.5 and c<5.3

The plot obtained is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 16

The obtained inequality is c≥−1.5 and c<5.3 and the plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 17

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5 Page 52 Exercise 6 Answer

Given: An inequality 6.4<x−3≤7.
To find The plot of the inequality.
Evaluate to get the answer.
Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 18

We solve for the inequality as follows
​4<x−3≤7                (Given )

4+3<x−3+3≤7+   ( Add 3 to both sides of the inequality )

7<x≤10

The plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 19

The obtained inequality is 7<x≤10 and the inequality obtained is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 20

 

Page 52 Exercise 7 Answer

Given: An inequality 15≥−5g≥−10.
To find The plot of the inequality.
Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 21

We solve for the inequality as follows

15≥−5g≥−10          (Given)

\(\frac{15}{-5} \geq \frac{-5 g}{-5} \geq \frac{-10}{-5}\)     (Divide both sides by −5)

-3≥g≥2                 ( Inequality symbol reverse)

−3≤g≤2

The plot obtained is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 22

 

The obtained inequality is −3≤g≤2 and the plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 23

Big Ideas Math Linear Inequalities Exercise 2.5 Help

Page 52 Exercise 8 Answer

Given: An inequality z+4<2 or −3z<−27.
To find The plot of the inequality.
Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 24

We solve for the inequality as follows

z+4<2​ or −3z<−27          (Given)

We separate this inequality to solve for z

1​.z+4<2z+4−4<2−4        (Subtract 4 on both sides)

z<−2\-3z<−27                   (Divide both sides by −3)

2.\(\frac{-3 z}{-3}<\frac{-27}{-3}\)

z<9                                   (Inequality reverses)

​z<−2 or z>9.

The plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 25

The obtained inequality is −2>z>9. and the plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 26

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5 Page 52 Exercise 9 Answer

Given: An inequality 2t+6<10 or−t+7≤2.
To find The plot of the inequality.
Evaluate to get the answer.
Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 27

We solve for the inequality as follows

2t+6<10 or −t+7≤2         (Given)
we separate this inequality to solve for t

​1. 2t+6<10                       (Subtract6 on both sides )

2t+6−6<10−6

2t<4

\(\frac{2 t}{2}<\frac{4}{2}\)

t<2

2.−t+7≤2

−t+7−7≤2−7              (Subtract both sides by 7)

−t≤−5

t≥5t<2 or t≥5.

The plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 28

The obtained inequality is t≥5t<2 or t≥5. and the plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 29

How To Solve Exercise 2.5 Big Ideas Math

Page 52 Exercise 10 Answer

Given: An inequality \(-8 \leq \frac{1}{3}(6 x+24) \leq 12\).
To find The plot of the inequality.
Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 30

Solve for the inequality as follows
\(-8 \leq \frac{1}{3}(6 x+24) \leq 12\)         (Given)

\(-8 \leq \frac{1}{3}(6 x)+\frac{1}{3}(24) \leq 12 \)     (Distribute and simplify)

-8≤2x+8≤24

−8−8≤2x+8−8≤24−8 &      ( Subtract 8 to both sides )

−16≤2x≤16

\(\frac{-16}{2} \leq \frac{2 x}{2} \leq \frac{16}{2}\)      (Divide both sides by 2)

-8≤x≤2

The plot is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 31

The obtained inequality is −8≤x≤2. The plot obtained is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.5 Solving compound Inequalities graph 32

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.5 Page 52 Exercise 11 Answer

Given: An inequality−60≤(h−50)≤60.
To find The range of values in which the machine operates.
Evaluate to get the answer.

Let us consider and solve for h

​−60≤2(h−50)≤60            (Given)

−60≤2(h)+2(−50)≤60     (Distribute )

−60≤2h−100≤60

−60+100≤2h−100+100≤60+100 (Add 100 to both sides of the inequality )

40≤2h≤160

\(\frac{40}{2} \leq \frac{2 h}{2} \leq \frac{160}{2}\)   (Divide both sides by 2)

20≤h≤80.

The obtained inequality after the evaluation is 20≤h≤80.

Big Ideas Math Student Journal Exercise 2.5 Examples

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.4

Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities

 


Page 43 Essential Question Answer

Clear parenthesis on both sides of the inequality and collect-like terms.

Adding or subtracting terms so the variable is on one side and the constant is on the other side of the inequality sign.

And by multiplying and dividing by whatever constants are attached to the variable.

Also, remember to change the direction of the inequality if you multiply or divide by a negative number.

We can solve a multi-step inequality by solving the equation so that the variable are on one side and the constant are on the other side and then solving the equation and also to change the direction of the inequality if you multiply or divide by a negative number.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Page 43 Exploration 1 Answer

Given: The inequality is  2x+3≤x+5
To Find  Solve the multi-step inequality.
Evaluate the question to get the answer

From the given we get
2x+3≤x+5
2x+3−3≤x+5−3
2x≤x+2
2x−x≤x+2−x
x≤2

After solving 2x+3≤x+5 we get x≤2.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.4

Given: The inequality is −2x+3>x+9
To Find Solve the multi-step inequality.
Evaluate the question to get the answer.

From the given, we get that,
−2x+3>x+9
−2x+3−9>x+9−9
−2x−6>x
−2x+2x−6>x+2x
−6>3x
\(\frac{-6}{3}>\frac{3 x}{3}\)
−2>x

After solving −2x+3>x+9 we get x<−2.

 

Given: The inequality is 27≥5x+4x
To Find: Solve the multi-step inequality.
Evaluate the question to get the answer.

From the given we get
27≥5x+4x
27≥ 9x
\(\frac{27}{9} \geq \frac{9 x}{9}\)
3≥x

After solving 27≥5x+4x we get x≤3.

 

Given: The inequality is−8x+2x−16<−5x+7x
To Find Solve the multi-step inequality.
Evaluate the question to get the answer.

From the given we get
−8x+2x−16<−5x+7x
−6x−16<2x
−6x−16+6x<2x+6x
−16<8x
\(\frac{-16}{8}<\frac{8 x}{8}\)
−2<x

After solving−8x+2x−16<−5x+7x we get x>−2.

 

Given: The inequality is  3(x−3)−5x>−3x−6
To Find  Solve the multi-step inequality.
Evaluate the question to get the answer.

From the given we get
3(x−3)−5x>−3x−6
3x−9−5x>−3x−6
−2x−9>−3x−6
−9>−x−6
−3>−x
3<x

After solving3(x−3)−5x>−3x−6 we get  x>3.

 

Given The inequality is −5x−6x≤8−8x−x
To Find: Solve the multi-step inequality.
Evaluate the question to get the answer.

From the given we get
−5x−6x≤8−8x−x
−11x≤8−9x
−11x+9x≤8−9x+9x
−2x≤8
x≥−4

After solving−5x−6x≤8−8x−x we get x≥−4.

Big Ideas Math Algebra 1 Chapter 2 Exercise 2.4 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.4 Page 44 Exercise 3 Answer

Given: The graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 1

To Find Write two different multi-step inequalities whose solutions are represented by the graph.
Evaluate the question to get the answer.

From the given graph we get the inequality to be
x<−1

Now, adding 3x to both sides we get
x+3x<−1+3x
4x<−1+3x

Again adding 2 to both sides we get
4x+2<−1+3x+2
4x+2<3x+1

The two multi-step inequalities whose solutions are represented by the graph are, 4x<−1+3x
and 4x+2<3x+1.

 

Page 46 Exercise 1 Answer

Given: The value is 3x−2<10.
The graph is

To solve the inequality and Graph the solution.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for x:

3x−2<10                ( Given)

3x−2+2<10+2​      (Add 2 to both sides)

3x<12                   (Simplify )

\(\frac{3 x}{3}<\frac{12}{3}\)       (Divide both sides by 3)

x<4

Plot x<4 as follows

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 3

The required solution is plot x<4 and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 4

Algebra 1 Student Journal Chapter 2 Exercise 2.4 Answers

Page 46 Exercise 2 Answer

Given: The value is 4a+8≥0.
The graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi- set up Inequalities graph 5

To find  To solve the inequality, Graph the solution.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for a:

​4a+8≥0                 (Given)

4a+8−8≥0−8        ( Subtract 8 from both sides )

4a≥−8

\(\frac{4 a}{4} \geq \frac{-8}{4}\)         ( Divide both sides by 4)

Plot a≥−2 as follows

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 6

The required solution is plot x≥−2 and the graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 7

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.4 Page 46 Exercise 3 Answer

Given: The value is \(2+\frac{b}{-3} \leq 3\).
The graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi- set up Inequalities graph 8

To solve the inequality, Graph the solution.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for b:

2+\(\frac{b}{-3} \leq 3\)        ( Given)

2+\(\frac{b}{-3}-2 \leq 3\)    (Subtract 2 to both sides )

\(\frac{b}{-3} \leq 1\)

\(\frac{b}{-3} \cdot(-3) \leq 1 \cdot(-3)\)  (Multiply both sides by −3)

b≤−3

b≥−3​                    ( Inequality symbol reverse)

b≥−3

Plot b≥−3 as follows

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 9

The required solution is Plot b≥−3 and the graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 10

 

Page 46 Exercise 4 Answer

 

Given : The value is −\(-\frac{c}{2}-6>-8\).The graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 11

To find To solve the inequality, Graph the solution.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for c:

\(-\frac{c}{2}-6>-8\)              (Given)

\(-\frac{c}{2}\) −6+6>−8+6  (Add 6 to both sides)

\(-\frac{c}{2}>-2\)                 (Inequality symbol reverses)

c<−4               (Multiply both sides by −2)

c<−4

​Plot  c <−4 as follows

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 12

The required solution is c<−4 graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 13

 

Page 46 Exercise 5 Answer

Given: The value is 8≤−4(d+1).
The graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 14

To solve the inequality, Graph the solution.
Evaluate to get the solution.

Let us consider the given value and simplify

​Solve for d :

​8≤−4(d+1)             (Given)

8≤−4d−4               ( Distribute )

8+4≤−4d−4+4     ( Add 4 to both sides)

12≤−4d

\(\frac{12}{-4} \leq \frac{-4 d}{-4}\)       (Divide both sides by −4)

−3≤d                  (Inequality symbol reverses )

−3≥d d≤−3

Plot  d≤−3 as follows

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 15

The required solution is plot d≤−3 and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 16

Chapter 2 Exercise 2.4 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.4 Page 47 Exercise 6 Answer

Given: The value is 5−2n>8−4n.
To find: To solve the inequality.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for n :

​5−2n>8−4n              (Given)

5−2n−5>8−4n−5     (Subtract 5 to both sides )

−2n>3−4n

−2n+4n>3−4n+4n     (Add 4n to both sides )

2n>3

\(\frac{2 n}{2}>\frac{3}{2}\)

\(n>\frac{3}{2}\)              (Divide both sides by 2)

Plot \(n>\frac{3}{2}\) as follows

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 17

The required solution is\(n>\frac{3}{2}\) and graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.4 Solving Multi-setup Inequalities graph 18

 

Page 47 Exercise 7 Answer

Given: The value is  6h−18<6h+1
To solve the inequality.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for h:

​6h−18<6h+1​               ( Given)

6h−18+18<6h+1+18​  (Add 18 to both sides)

6h<6h+19

6h−6h<6h+19−6h​     (Subtract oh to both sides )

0<19
There is no solution to this inequality

The result is 0<19 there is no solution to this inequality.

 

Page 47 Exercise 8 Answer

Given: The value is 3p+4≥−4p+25
To solve the inequality.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for p:

​3p+4≥−4p+25                (Given)

3p+4+4p≥−4p+25+4p  (Add 4p to both sides)

7p+4≥25

7p+4−4≥25−4            ( Subtract 4 to both sides )

7p≥21

\(\frac{7 p}{7} \geq \frac{21}{7}\)           (Divide both sides by 7)

p≥3

Hence the solution of the given inequality 3p+4≥−4p+25 is p≥3.

Big Ideas Math Chapter 2 Exercise 2.4 Walkthrough

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.4 Page 47 Exercise 9 Answer

Given: The value is 7j−4j+6<−2+3j.
To solve the inequality.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for j :

​7j−4j+6<−2+3j         ( Given)

3j+6<−2+3j              (Simplify)

3j+6−6<−2+3j−6    ( Subtract 6 to both sides)

3j<3j−8

3j−3j<3j−8−3j         (Subtract 3j to both sides )

0<−8
There is no solution to this inequality

The required solution is 0<−8, there is no solution to the inequality.

 

Page 47 Exercise 10 Answer

Given : The value is 12\(12\left(\frac{1}{4} w+3\right) \leq 3(w-4)\).
To find: To solve the inequality.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for w:

​12\(12\left(\frac{1}{4} w+3\right) \leq 3(w-4)\)  (Given)

\(\frac{12}{4} w\)+12⋅3≤3w−3⋅4​      (Distribute)

3w+36≤3w−12                                    (Simplify)

3w+36−36≤3w−12−36​                       (Subtract 36 to both sides)

3w≤3w−48

3w−3w≤3w−48−3w ​                           (Subtract 3w to both sides )

0≤−48

There is no solution to this inequality

The required solution is 0≤−48, there is no solution to the inequality.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Page 47 Exercise 11 Answer

Given : The value is k(4r−5)≥−12r−9
To find the value of k.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for k:

To have all real numbers as a solution for the inequality, we must find a value of k that is independent of r and the other constants.

We solve for as
k(4r−5)≥−12r−9
Given remove constants and inequality symbol
​k4r=−12r
\(\frac{k 4 r}{4 r}=\frac{-12 r}{4 r}\) ( Divide both sides by 4r)

  
Given: k(4r−5)≥−12r−9  ,k=−3

​k(4r−5)≥−12r−9    ​     (Given )

−3(4r−5)≥−12r−9      (k=−3 distribute)

−3⋅4r−3⋅(−5)≥−12r−9

Test 1:

−12r+15≥−12r−9          (Set r=0)

−12(0)+15≥−12(0)−9    (Simplify)

15≥−9                           (This is true )

Test 2:

−12r+15≥−12r−9          (Set r=1)

−12(1)+15≥−12(1)−9    (Simplify)

3≥−21                          (This is also true)

Hence, if k=−3, the solution of r is the set of all real numbers

The required solution is k=−3 the solution of r is the set of all real numbers.

Page 47 Exercise 12 Answer

Given: The value is 2kx−3k<2x+4+3kx.
To find: To find the value of k.
Evaluate to get the solution.

Let us consider the given value and simplify

Solve for k:

Given, 3kx−3k<2x+4+3kx,

find the value of k such that this inequality has no solution. We only consider the values that are dependent on x, such that
​3​kx−3k<2x+4+3kx             (Given)

2kx=2x+3kx                  ( Constants and inequality symbol are removed)

\(\frac{2 k}{x}\)=\(\frac{2 x}{x}+\frac{3 k x}{x}\) (Divide all terms by x)

2k=2+3k

2k−3k=2+3k−3k​

−k=2

−k(−1)=2(−1)​

k=−3                    ( Subtract 3k to both sides)

Checking:

​3kx−3k<2x+4+3kx               ​        ( Given )

3(−2)x−3(−2)<2x+4+3(−2)x        ( Set k=−2)

Test:1

3(−2)x−3(−2)<2x+4+3(−2)x          (Set x=0)

3(−2)(0)−3(−2)<2(0)+4+3(−2)(0)  ( Simplify )

0+6<0+4+0

6<4                                              ( This is not true )

Test 2:

3(−2)x−3(−2)<2x+4+3(−2)x             (Set x=1)

3(−2)(1)−3(−2)<2(1)+4+3(−2)(1)      ( Simplify )

−6+6<2+4+−6

0<0                                                  ( This is not true )

Hence, if k=−2, any value of r is not a solution to this inequality

The required solution is k=−2 which makes the inequality 2kx−3k<2x+4+3kx to have no solution.

Algebra 1 Exercise 2.4 Explanation Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1

Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities

Page 28 Essential Question Answer

When two values are compared, an inequality shows whether one is less than, greater than, or simply not equal to the other is known as inequality.

Inequality is the relationship between two amounts.

Here the amounts are not the same.

For example, the speed limit in highways is less than or equal to 65 miles per hour.

Inequality in real life may be stated as the speed limit in highways is less than or equal to 65 miles per hour.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Page 28 Exercise 1 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 1

To find: The temperature t in Sweden is at least−10∘ C.
Evaluate to get the final answer.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1

If t is temperature it will be represented as the symbol ≥.
The inequality will be  t ≥−10.
Graphically it will be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 2

The graphical representation of the inequality t≥−10 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 3

 

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 4

To find: The elevation e of Alabama is at most 2047 feet.
Evaluate to get the final answer.

If e is elevation of Alabama it will be represented as the symbol ≤.
The inequality will be: e≤2047.
Graphically it will be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 5

The graphical representation of the inequality e≤2047 is

Big Ideas Math Algebra 1 Chapter 2 Exercise 2.1 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1 Page 28 Exercise 2 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 6

To find The values of x.
Evaluate to get the final answer.

In the graph, we can see that x is at least 1.
Or x is greater than or equal to 1
We can depict it as  x≥1.
Graphically we can represent it as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 7

In words, we can write x is less than or equal to 1 graphically x≥1 can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 8

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 9

To find The values of x.
Evaluate to get the final answer.

In the graph, we can see that x is always greater than 1.
We can depict it as  x>1.
Graphically we can represent it as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 10

In  words, all the values of x that satisfy the given situation is x
is always greater than 1 graphically  x>1 c an be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 11

 

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 12

To find The values of x.
Evaluate to get the final answer.

In the graph, we can see that x is always greater than or equal to 1.
We can depict it as x≤1.
Graphically we can represent it as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 13

 

In words, we can write that x is less than or equal to 1 graphically x≤1
Can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 14

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 15

To find The values of x.
Evaluate to get the final answer.

In the graph, we can see that x is always greater than 1.
We can depict it as x<1.
Graphically we can represent it as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 16

Graphically x<1 is represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 17

Solving Linear Inequalities Exercise 2.1 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1 Page 29 Exercise 3 Answer

When two values are compared, an inequality shows whether one is less than, greater than, or simply not equal to the other is known as inequality.

Inequality is the relationship between two amounts.
Here the amounts are not the same.
For example, the speed limit in highways is less than or equal to 65 miles per hour.

Inequality in real life may be stated as the speed limit in highways is less than or equal to 65 miles per hour.

 

Page 31 Exercise 3 Answer

Let the number be q.
The inequality to represent seven is less than or equal to the difference of a number and 6 is  7<q−6.

The representation of seven is less than or equal to the difference of a number and 6 is  7<q−6.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1 Page 31 Exercise 4 Answer

Let the number be u.
The inequality to represent the sum of a number and 14 is more than 6 is  u+14>6.

The representation of the sum of a number and 14 is more than 6 is u+14>6.

 

Page 31 Exercise 5 Answer

Given: The inequality d−7<12.
To find The value=19.
Evaluate to get the final answer.

We will substitute d=12 in the given inequality.
Then, it will be evaluated as  19−7<12 12<12
So,d=12 is not a solution for the given equality.

The value d=12 is not a solution for the given inequality.

 

Page 31 Exercise 6 Answer

Given: The inequality 9≥3n+6.
To find The value n=1.
Evaluate to get the final answer.

We will substitute n=1 in the given inequality.
Then, it will be evaluated as:​

9≥3n+6
9≥3(1)+6
9≥3+6
9≥9

So,n=1 is a solution for the given equality.

The value n=1 is a solution for the given equality.

Algebra 1 Student Journal Chapter 2 Exercise 2.1 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1 Page 31 Exercise 8 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 17

To find The inequality in the graph.
Evaluate to get the final answer.

 

The inequality x≤4 will be plotted in the graph.
It will be shown as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 18

The inequality x≤4 will be marked in the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 19

 

Page 32 Exercise 9 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 20

To find The inequality in the graph.
Evaluate to get the final answer.

The inequality x>−1 will be plotted in the graph.
It will be shown as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 21

The inequality x>−1 will be marked in the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 22

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1 Page 32 Exercise 10 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 23

To find The inequality in the graph.
Evaluate to get the final answer.

 

The inequality x<1 will be plotted in the graph.
It will be shown as:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 25

The inequality x<1 will be marked in the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 26

Page 32 Exercise 11 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 27

To find: The inequality marked in the graph.
Evaluate to get the final answer.

The dot is marked at 1 and the graph is shifting towards right.
It means that x is greater than 1.
The inequality to represent the graph is x>1.

We will observe the graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 28

The graph states thatx>1.

Big Ideas Math Linear Inequalities Exercise 2.1 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.1 Page 32 Exercise 13 Answer

Given: The graph
To find: The inequality marked in the graph.
Evaluate to get the final answer.

The dot is marked at−3 and the graph is shifting toward  right.
It means that x is greater than equal to−3.
The inequality to represent the graph is x≥−3.

We will observe the graph.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 29

The graph states that x≥−3.

 

Page 32 Exercise 14 Answer

Here since the dot is at 3 and the dot is filled in.
Which could be equal to 3 and the graph is filled into the left.
Therefore x is less than or equal to 3.

An inequality that represents the given graph is x≤3.

Chapter 2 Exercise 2.1 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2

Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities

Page 33 Essential Question Answer

If we want to add or subtract from one side of the equation, we must perform the same operation to the other side of the equation.

This ensures that the inequality is still true. When solving inequalities by subtracting, our goal is to have the variable on its own.

Adding or subtracting will not change the directions of the < or > than signs.

For using addition or subtraction to solve an inequality we need to perform the same action on both sides of the equation.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Page 33 Exploration 1 Answer

Given:  \(P=\frac{8.4 Y+100 C+330 T-200 N}{A}\) and Y is the total length of all completed passes (in Yards), C is Completed passes, T is passes resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes

To Find  Whether T<C is true or not.
Evaluate the question to get the answer.

From the given we get,
A=C+N+M…….(1)

C=T+Nt……..(2)

Now, using equation (2) we get,
T<C
T<T+Nt
C>T
Thus, T<C.
This equation is valid given that all the variables are greater than or equal to zero.

Hence, T<C is true because this equation is valid given that all the variables are greater than or equal to zero.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2

Given:  \(P=\frac{8.4 Y+100 C+33 U I^{\prime}-200 N}{A}\)and Y is the total length of all completed passes (in Yards), C is Completed passes,T is passed resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes.

To Find  Whether C+N≤A is true or not.
Evaluate the question to get the answer.

From the given we get,
A=C+N+M ……..(1)

C=T+Nt……..(2)

Now, using equation (1) we get

C+N≤A
C+N≤C+N+M
A≥C+N
Thus, C+N≤A.
This equation is valid given that all the variables are greater than or equal to zero.

This equation is valid given that all the variables are greater than or equal to zero C+N≤A
is true.

 

Given:   \(P=\frac{8.4 Y+100 C+33 U I^{\prime}-200 N}{A}\) and Y is the total length of all completed passes (in Yards), C is Completed passes, T is passes resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes.
To Find  Is N<A true.
Evaluate the question to get the answer.

From the given we get,
A=C+N+M…….(1)

C=T+Nt………(2)

Now, using equation(1) we get
N<A
N<C+N+M
A>N
Thus, N<A.
This equation is valid given that all the variables are greater than or equal to zero.

This equation is valid given that all the variables are greater than or equal to zero N<A is true.

Big Ideas Math Algebra 1 Chapter 2 Exercise 2.2 Solution

Given: \(P=\frac{8.4 Y+100 C+33 U I^{\prime}-200 N}{A}\) and Y is the total length of all completed passes (in Yards), C is Completed passes, T is passes resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes.

To Find  Is A−C≥M true.
Evaluate the question to get the answer.

From the given we get,
A=C+N+M……..(1)

C=T+Nt………(2)

Now, using equation(1) we get

A−C≥M
(C+N+M)−C≥M
N+M≥M
Thus, A−C≥M.
This equation is valid given that all the variables are greater than or equal to zero.

A−C ≥M is true.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 34 Exercise 3 Answer

To determine if an inequality is legitimate, we utilize addition and subtraction.

In addition, we solve inequalities using addition and subtraction in the same way as we solve linear equations.

We use addition and subtraction to solve an inequality the same way we solve an equality.

 

Page 34 Exercise 4 Answer

To Find Solve the inequality x+3<4
Evaluate the question to get the answer.

Consider the following inequality
​x+3<4
=x+3−3<4−3
=x<1

After solving x+3<4 we get x<1.

 

To Find  Solve the inequality x−3≥5
Evaluate the question to get the answer.

Consider the inequality
​x−3≥5
=x−3+3≥5+3
=x≥8

After solving x−3≥5 we get, x≥8.

 

To Find  Solve the inequality 4>x−2
Evaluate the question to get the answer.

From the given we get,
​4<x−2
4+2<x
x>6

After solving 4>x−2 we get x>6.

 

To Find  Solve the inequality −2≤x+1
Evaluate the question to get the answer.

From the given we get,
−2≤x+1
−2−1≤x+1−1
x≥−3

After solving −2≤x+1 we get x≥−3.

Solving Linear Inequalities Exercise 2.2 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 36 Exercise 1 Answer

To Find  Solve the inequality x−3<−4 and graph the solution.
Evaluate the question to get the answer.

From the given, we get that,
x−3<−4
x−3+3<−4+3
x<−1

Now, graphing the solution we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 1

After solving x−3<−4 we get x<−1 and the graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 2

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 36 Exercise 2 Answer

To Find Solve the inequality −3>−3+h and graph the solution.
Evaluate the question to get the answer.

From the given we get
−3>−3+h
−3+3>−3+h+3
0>h

Now, graphing the solution we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 3

After solving −3>−3+h we get h<0 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 4

Algebra 1 Student Journal Chapter 2 Exercise 2.2 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 37 Exercise 3 Answer

To Find  Solve the inequality s−(−1)≥2 and graph the solution.
Evaluate the question to get the answer.

From the given we get,
s−(−1)≥2
s+1≥2
s+1−1≥2−1
s≥1

Now, graphing the solution we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 5

After solving s−(−1)≥2 we get s≥1 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 6

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 37 Exercise 5 Answer

To Find  Solve the inequality 12≤4c−3c+10 and graph the solution.
Evaluate the question to get the answer.

From the given we get,
12≤4c−3c+10
12≤c+10
12−10≤c+10−10
2≤c

Now, graphing the solution we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 7

After solving 12≤4c−3c+10 we get c≥2 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 8

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 37 Exercise 6 Answer

To Find Solve the inequality 15−7p+8p>15−2
and graph the solution.
Evaluate the question to get the answer.

From the given we get,
15−7p+8p>15−2
15+p>13
15+p−15>13−15
p>−2

Now, graphing the solution we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 9

After solving 15−7p+8p>15−2 we get p>−2 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.1 Writing and Graphing Inequalities graph 10

Big Ideas Math Linear Inequalities Exercise 2.2 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.2 Page 37 Exercise 7 Answer

Given: The amount of money to be spend is $15 and amount of groceries in the cart is$12.25
To Find  Write an inequality that represents how much more money m that can be spend.
Evaluate the question to get the answer.

Let ,S=15 , C=12.25

Now, from the given we get the inequality as
S≥C+m……(1)

S≥C+m is the inequality that represents how much more money m can be spend.

 

Given: The amount of money to be spend is $15 and the amount of groceries in the cart is$12.25
To Find: Solve the inequality.
Evaluate the question to get the answer.

Given:  $15 to spend on groceries and $12.25 worth of groceries already in your cart.
Let, S=15 , C=12.25

Substituting the values in we get
S≥C+m
15≥12.25+m
15−12.25≥12.25+m−12.25
2.75≥m

After solving the inequality we get m≤2.75.

Chapter 2 Exercise 2.2 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3

Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities

Page 38 Essential Question Answer

Solving inequalities is very similar to solving equations.

It just that when we divide we need to reverse the symbol on both sides of the equation by a negative number.

We can solve an inequality by using division just the way we solve an equation just we need to reverse the symbols of both the sides by a negative number.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Page 38 Exercise 1 Answer

To Find Complete the table. Decide which graph represents the solution of the inequality 6<3x
and write a solution of the inequality.
Evaluate the question to get the answer.

From the given we get,
6<3x
\(\frac{6}{3}<\frac{3 x}{3}\)
2<x

Now, the graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 1

Now, using the graph we get the table as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter2.3 Solving Inequalities Using Multiplication or Division Table 1

The graph that represents the solution of inequality 6<3x is

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 2

And the table is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter2.3 Solving Inequalities Using Multiplication or Division table 2

 

Given: The inequalities are,2x<4, 3≥3x, 2x<8, 6≥3x
To Find Use a table to solve each inequality. Then write a rule that describes how to use division to solve the inequalities.
Evaluate the question to get the answer.

From the given, we get that
(i): 2x<4
x<2

(ii): ​3≥3x
x≤1

(iii): ​2x<8
x<4

(iv): ​6≥3x
x≤2

Now, from the solution above we get the table as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter2.3 Solving Inequalities Using Multiplication or Division table 3

One of those rules is called the division property of inequality, and it basically says that if you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.

The tables for solving the given inequalities are

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving Inequalities Using Multiplication or Division table 4

One of those rules is called the division property of inequality, and it basically says that if you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.

Big Ideas Math Algebra 1 Chapter 2 Exercise 2.3 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3 Page 39 Exercise 2 Answer

To Find  Complete the table. Decide which graph represents the solution of the inequality 6<−3x
and write solutions for the inequalities.
Evaluate the question to get the answer.

From the given we get,
6<−3x
\(\frac{6}{-3}<\frac{-3 x}{-3}\)
−2>x

Now, plotting the solution we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 3

Now, using the result we complete the table as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving Inequalities Using Multiplication or Division table 5

After solving 6<−3x we getx<−2 and the graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 4

And the table is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving Inequalities Using Multiplication or Division table 6

Given: The inequalities are−2x<4,3≥−3x,−2x<8 & 6≥−3x
To Find  Use a table to solve each inequality. Then write a rule that describes how to use division to solve the inequalities.
Evaluate the question to get the answer.

From the given we get
(i): −2x<4
x>−2

(ii): 3≥−3x
x≥−1

(iii): −2x<8
x>−4

(iv):  6≥−3x
x>−2

Using the result we get the table as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving Inequalities Using Multiplication or Division table 7

One rule is called the division property of inequality, and it basically says that if you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number. And change the sign of inequality if the number is negative.

 

The tables for solving the inequality are

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving Inequalities Using Multiplication or Division table 8

One rule is called the division property of inequality, and it basically says that if you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number. And change the sign of inequality if the number is negative.

Solving Linear Inequalities Exercise 2.3 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3 Page 39 Exercise 3 Answer

Solving inequalities is very similar to solving equations.
But, when we multiply/divide by a negative number the sign changes direction >→< and vice versa.

Solving inequalities is very similar to solving equations but, when we multiply/divide by a negative number the sign changes direction>→< and vice versa

 

Page 39 Exercise 4 Answer

Given: The inequality is7x<−21
To Find  Solve using the rule in Exercises (1) & (2)
Evaluate the question to get the answer.

Using the rule in Exercises (1) & (2)
in the given inequality we get
​7x​<−21
\(\frac{7 x}{7}<\frac{-21}{7}\)
x<−3

After solving the inequality 7x<−21 we get x<−3.

 

Given: The inequality is 12≤4x
To Find: Solve using the Exercise (1) & (2)
Evaluate the question to get the answer.

Using the rule in Exercises (1) & (2) we get
12≤4x
\(\frac{12}{4} \leq \frac{4 x}{4}\)
3≤x

After solving 12≤4x using Exercises  (1) & (2) we get x≥3.

 

Given: The inequality is 10<−5x
To Find  Solve using the Exercise (1) & (2)
Evaluate the question to get the answer.

Using the  Exercises  in (1) & (2) we get,
10<−5x
−2>x

After solving 10<−5x using the Exercise in (1) & (2) we get x<−2.

 

Given: The inequality is −3x≤0
To Find: Solve using the Exercise  in (1) & (2)
Evaluate the question to get the answer.

Using the Exercises  in (1) & (2) we get
−3x≤0
x≥0

After solving −3x≤0 using the Exercise in  (1) & (2) we get x≥0.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3 Page 41 Exercise 1 Answer

To Find  Solve 16x<−30 and plot the graph.
Evaluate the question to get the answer.

From the given we get
6x<−30
\(\frac{6 x}{6}<\frac{-30}{6}\)
x<−5

Now, plotting the result we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 5

After solving 16x<−30 we get x<−5 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 6

 

Page 41 Exercise 2 Answer

To Find  Solve 48≤16f  and graph the result.
Evaluate the question to get the answer.

From the given we get
48≤16f
3≤f.

Now, plotting the result we get the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 7

After solving 48≤16f we get f≥3 and the graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 8

Algebra 1 Student Journal Chapter 2 Exercise 2.3 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3 Page 41 Exercise 3 Answer

To Find Solve\(\frac{-6}{7} \leq \frac{3}{7} f\) and graph the result.
Evaluate the question to get the answer.

From the given, we get that
\(\frac{-6}{7} \leq \frac{3}{7} f\)
−2≤f

Now, plotting the result we get the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 9

After solving\(\frac{-6}{7} \leq \frac{3}{7} f\) we get f≥−2 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 10

 

Page 41 Exercise 4 Answer

To Find Solve −4m≥−16 and graph the result.
Evaluate the question to get the answer.

From the given we get
−4m≥−16
m≤4

Now, plotting the result we get

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 11

After solving −4m≥−16 we get m≤4 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 12

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3 Page 41 Exercise 5 Answer

To Find  Solve \(\frac{x}{-6}>\frac{1}{3}\) and graph the result.
Evaluate the question to get the answer.

From the given we get
\(\frac{x}{-6}>\frac{1}{3}\)
x<−2

Now, plotting the result we get the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 13

After solving \(\frac{x}{-6}>\frac{1}{3}\) and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 14

Page 42 Exercise 6 Answer

To Find  Solve\(1 \leq-\frac{1}{4} y\) and graph the result.
Evaluate the question to get the answer.

From the given we get
\(1 \leq-\frac{1}{4} y\)
(1)(−4)≤\(-\frac{1}{4} y(-4)\)
−4≥y

Now, plotting the result we get the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 15

After solving 1≤−1\(1 \leq-\frac{1}{4} y\) we get y ≤−4 and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 16

Big Ideas Math Linear Inequalities Exercise 2.3 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2.3 Page 42 Exercise 7 Answer

To Find Solve\(\frac{-2}{3}<-4 x\) and graph the result.
Evaluate the question to get the answer.

From the given we get
\(\frac{-2}{3}<-4 x\)
\(x<\frac{1}{6}\)

Now. plotting the result we get the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 17

After solving \(\frac{-2}{3}<-4 x\) we get \(x<\frac{1}{6}\) and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 18

 

Page 42 Exercise 8 Answer

To Find  Solve\(\frac{-4}{5} x \geq-2\)and graph the result.
Evaluate the question to get the answer.

From the given we get
\(\frac{-4}{5} x \geq-2\)
\(x \leq \frac{5}{2}\)

Now, plotting the result we get the graph as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 19

After solving\(\frac{-4}{5} x \geq-2\) we get \(x \leq \frac{5}{2}\) and the required graph is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2.3 Solving inequalities using multiplication or division graph 20

Chapter 2 Exercise 2.3 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 1 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 1

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

To find The integer for∣−2∣.
Evaluate to get the final answer.

The absolute value is|−2| is 2.
So, it can be shown as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 2

 

The numerical value is|−2|=2.
Graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 3

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2

Big Ideas Math Algebra 1 Chapter 2 Exercise 2 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 2 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 4

 

To find The integer for−3+|−3|.
Evaluate to get the final answer.

 

The absolute value of −3 is 3.
Then, it will be shown as
−3+∣−3∣
=−3+3
=0

Graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 5

 

The numerical value is−3+|−3|=0.
Graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 6

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 3 Answer

Given: The graph

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 7

 

To find The integer for−1−|−4|.
Evaluate to get the final answer.

 

The absolute value of−4 is 4.
Then it will be shown as
​−1−|−4|
=−1−4
=−5

​Graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 8

 

The numerical value is=−1−|−4|=−5.
Graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 9

Solving Linear Inequalities Exercise 2 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 4 Answer

Given: The graph
To find The integer for 2+|2|.
Evaluate to get the final answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 10

 

The absolute value of 2 is 2.
Then it will be shown as:
​2+|−2|
=2+2
=4

So, graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 11

The numerical value is2+|−2|=4.
Graphically it can be represented as

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 2 Solving Linear Inequalities graph 12

Algebra 1 Student Journal Chapter 2 Exercise 2 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 6 Answer

We know that 3 is greater than−2.
So, the symbol used is<.
The numeric value is−2<3.

The sentence will be completed as −2<3.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 7 Answer

We know that−4 is greater than−7.
So, the symbol used is>.
The numeric value is−4>−7.

The sentence will be completed as−4>−7.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 8 Answer

We know that−5 is greater than−8.
So, the symbol used is <.
The numeric value −8<−5.

The sentence will be completed as −8<−5.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 9 Answer

The absolute value of|−5| is 5
So, the symbol used is=
The numeric is |-5 |=5.

The sentence will be completed as |−5|=5.

Big Ideas Math Linear Inequalities Exercise 2 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 10 Answer

The absolute value of|−6| is 6.
We know that −6 is greater than−7.
So, the symbol used is<.
The numeric value is−7<|−6|.

The sentence will be completed as −7<|−6|.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 11 Answer

We know that a is greater than b.
Here, we have the negative values.
So,−b is greater than−a.
It will be shown as  −a<−b.

In the statement, it is found that −a<−b.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities Exercise 2 Page 27 Exercise 12 Answer

We know that the number will always start from the left.

So,b is greater than a.

If a and b are negative numbers
.
Then, it will be shown as:|−a|>|−b|.

If a and b are positive numbers.

It will be depicted as  |−a|<|−b|.

The numbers a and b are negative numbers.
Then,|−a|>|−b|.
The numbers a and b are positive numbers.
Then,|−a|<|−b|.

Chapter 2 Exercise 2 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 22 Exercise 1 Answer

Given :

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 1

To write: The formula for area A of a parallelogram.
Evaluate to get the answer.

The formula of the area A of the parallelogram is:
A=bh ………(1)

The formula of the area A of the parallelogram is A=bh.

Given :

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 2

 

Using Eq(1) we can solve for  b:
A=bh (parallelogram formula)

30 in2 =b⋅(5in)plug−in given values
b.(5in)&=30in2                       (isolate b, divide by 5)
b.(5in)/5in =30in2\5in           (solve for b)
b=6in.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

The required solution is b=6in.

Given :

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 3

To find The value of b without substituting values in the formula.
Evaluate to get the answer.

Using Eq(1) we can solve for b:

A =bh         (parallelogram formula)
bh=A          (isolate b, divide by h)
bh/h=A/h   (solve for b)
b=A/h.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5

For part b, we plug in the given values first before we can solve for b, on the other hand, for part c, the result is just a formula for b using Eq(1).

The main difference between part b and part c is that we were able to get the value of b in part b however, for part c, we only obtained the formula for solving b.

The main difference between part b and part c is that we were able to get the value of b in part b, however, for part c we only obtained the formula for solving b.

Big Ideas Math Algebra 1 Chapter 1 Exercise 1.5 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 23  Exercise 2 Answer

 

Given :

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 4

 

To find The area A of a trapezoid.
Evaluate to get the answer.

 

The area of the trapezoid is given by : A=(b1+b2/2)h

Solve for h:

​A=(b1+b2/2)h​ formula
h=A(b​1+b​2/2)
h=63/(8)+(10)/2
h=63/18/2
h=63/9
h=7cm.

 

The required solution for area A of a trapezoid is 7cm.

 

Given :

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 5

 

To find Circumference C of circle.
Evaluate to get the answer.

 

The circumference of the circle is given by: C=2πr

Solve for r:

​C =2πr            (formula)
r=C/2π           (isolate r)
r=24π/2π      (solve for r)
r=12 ft

The required solution for Circumference C of circle is 12ft.

Given :

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 6

To find Volume V of a rectangular prism.
Evaluate to get the answer.

 

The volume of a rectangular prism is given by: V=Bh

Solve for h:

​V=Bh           (formula)
h=V/B         (isolate h)
h=75/15      (solve for h)
h=5yd2


The required solution for volume of a rectangular prism is 5yd2.

 

Given :

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .5 Extra practice image 7

 

To find Volume V of the cone.
Evaluate to get the answer.

The volume of a cone is given by: V=1/3πr2  h

Solve for h:

​V=1/3πr2h​         (formula)
h=3V/πr2            (isolate h)
B=πr2
h=3.24/12π       (solve for h)
h=6cm.

The required solution for volume of a cone is 6cm.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 23 Exercise 3 Answer

 

You can convert units easily and accurately with one simple rule:

Measurement by a carefully chosen form of the number 1.

Numbers of units treated exactly the same as coefficients with variables.

we can multiply anything by 1 and not change its value.

 

If you had a sample of a substance with a mass of 0.0034 grams, and you wanted to express that mass in mg, you could use the following dimensional analysis.

⇒ 1m gm=10-3 gm

∴ ⇒ 0.0034gm×1000mg/ 1g

=3.4g.

Any formula is made of dependent variables and independent variables. Suppose, you are given a measurement of one variable, you can rearrange the formula using operations such as addition subtraction, multiplication, division, and other properties of equality and solve for the unknown measurement.

Solving Linear Equations Exercise 1.5 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 1 Answer

 

Given: The expression is​ y−2x=15
Evaluate to get the answer.

On solving the given equation :

Solve for y:

​y−2x =15​ given
y−2x+2x=15+2x     (add 2x to both sides, isolate y)
y−0=15+2x
y=15+2x

The required solution is y=15+2x.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 2 Answer

 

Given: Expression is 4x+y=2.
Evaluate to get the answer.

 

On solving the given equation :

Solve for y:

​4x+y=2 given
4x+y−4x=2−4x          (subtract 4x to both sides, isolate y)
y+4x−4x=2−4x
y−0=2−4x
y=2−4x

The required solution is y=2−4x.

Algebra 1 Student Journal Chapter 1 Exercise 1.5 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 3  Answer

 

Given: The expression is​ 5x−2 =8+5y.
Evaluate to get the answer.

On solving the given equation :

Solve for y:

​5x−2 =8+5y given ,
5x−2−8=8+5y−8         (subtract both sides by 8, isolate 5y)
5x−10=5y                    (divide both sides by 5)
5y/5= 5x−10/5            (simplify)
y=5x/5-10/5
y=x-2

The required solution is y=x−2.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 4 Answer

 

Given: The expression is ​y+x=11.
Evaluate to get the answer.

On solving the given equation :

Solve for y:

​y+x=11 given,
y+x−x=11−x       (subtract x on both sides, isolate y)
y-0=11−x
y=11-x

The required solution is y=11−x.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 5 Answer

 

Given: The expression is 3x−y=−4
Evaluate to get the answer.

On solving the given equation :

Solve for y:

​3x−y=−4 given
3x−y−3x=−4−3x             (subtract 3x to both sides, isolate y).
(−1)⋅−y=(−1)⋅(−4−3x)     (multiply both sides by−1)
y=4+3x

 

The required solution is y=4+3x.

Big Ideas Math Linear Equations Exercise 1.5 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 6  Answer

 

Given: The expression is  3x+1=7−4y.
Evaluate to get the answer.

On solving the given equation:

Solve for y:

3x+1 =7−4y given
3x+1−7=7−4y−7        (subtract 7 to both sides, isolate y)
3x−6/−4=−4y/−4        (divide both sides by−4)
−3x−6/4=y
y=−3x−6/4

 

The required solution is y=−3x−6/4.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page  25 Exercise 7  Answer

 

Given: The equation y=10x−4x.
To find The value of x.
Evaluate to get the final answer.

 

We need to isolate the variable x from other variable y and we get the result.

The value of x can be evaluated as:​
y=10x−4x
y=(10−4)x
y=6x

The value of x in the equation y=10x−4x is:
y=6x.

 


Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25  Exercise 8  Answer

 

Given: The equation q=3x+9xz.
To find The value of x.
Evaluate to get the final answer.

 

We need to isolate the variable x from other variables q,z and we get the result.

The value of x can be evaluated as:​
q=3x+9xz
q=x(3+9z)
x=q/3+9z

The value of in the equation q =3x+9xz is:
x=q/3+9z.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 9 Answer

 

Given: The equation r=4+7x−sx.
To find The value of x.
Evaluate to get the final answer.

We need to isolate the variable x from other variables, s and we get the result.

The value of x can be evaluated as:​
r=4+7x−sx
r−4=7x−sx
r−4=(7−s)x
x=7−s/r−4

The value of x in the equation r=4+7x−sx is:
x=7−s/r−4.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 10 Answer

 

Given: The equation y+4x=10x−6.
To find The value of x.
Evaluate to get the final answer.

We need to isolate the variable from other variables, x  and we get the result.

The value of x can be evaluated as:
​y+4x=10x−6
10x−4x=y+6
6x=y+6
x=y+6/6

 

​The value of x in the equation  y+4x=10x−6 is:
x=y+6/6.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 11 Answer

 

Given: The equation 4g+r=2r−2x.
To find The value of x.
Evaluate to get the final answer.

We need to isolate the variable x from other variables g,r and we get the result.

The value of x can be evaluated as:​
4g+r=2r−2x
2x=2r−r−4g
2x=r−4g
x=r−4g/2

The value of x in the equation 4g+r=2r−2x is:
x=r−4g/2.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 12 Answer

 

Given: The equation 3z+8=12+3x−z.
To find The value of x.
Evaluate to get the final answer.

We need to isolate the variable x from other variables z and we get the result.

The value of x can be evaluated as:​
3z+8=12+3x−z
3x=3z+z+8−12
3x=4z−6
x=4z−6/3

 

The value of x in the equation 3z+8=12+3x−z is:
x=4z−6/3.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 13 Answer

 

Given: The equation A=1/2 bh.
To find The value of b.
Evaluate to get the final answer.


We need to isolate the variable b from other variables h, A and we get the result.

The value of b can be evaluated as:
​A=1/2 bh
2A=bh
b=2A/h

 

​The value of b found after evaluation is 2A/h.

Chapter 1 Exercise 1.5 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 14  Answer

 

Given: The equation V=1/3πr2  h.
To find The value of h.
Evaluate to get the final answer.

We need to isolate the variable h from other variables, V and we get the result.

The value of h can be evaluated as​ V=1/3πr2h
3V=πr2h
h=3V/π.r2

The value of h found after evaluating is h=3V/π.r2

 


Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 26 Exercise 15 Answer

 

Given: The equationI=V/R.
To find The value of R.
Evaluate to get the final answer.

 

We need to isolate the variable R from other variable, V and we get the result.
The value of R can be evaluated as​I=V/R
IR=V
R=V/I

The value of R found after evaluating is   V/I.

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 26 Exercise 16 Answer

 

Given: The equation PV=nRT.
To find The value of R.
Evaluate to get the final answer.

We need to isolate the variable R from other variables, P, V, and T and we get the result.
The value of R can be evaluated as   ​PV=nRT
R=PV/nT.

The value of R found after evaluating is  PV/nT.

Exercise 1.5 Big Ideas Math Algebra 1 Guide Informational

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 26 Exercise 17 Answer

 

Given: The equation A=P+Prt.
To find The value of r.
Evaluate to get the final answer.

We need to isolate the variable from other variables A, P, and t and we get the result.
The value of r can be evaluated as  ​A=P+Prt
Prt=A−P
rt=A−P/P
r=A−P/Pt


The value of r found after evaluating is  A−P/Pt.

 

Given: The amount after interest earned is $1080, and time is 2 years.
To find The annual interest rate.
Evaluate to get the final answer.

 

The interest for 2 years can be evaluated as:
$1080−1000=80.

 

The annual interest for 1 year is 80/2=40.
The annual interest rate is $40.

 

Given: The equation A=P+Prt.
To find The formula for P.
Evaluate to get the final answer.

We need to isolate the variable P from other variables, A,r, and t and we get the result.
The value of P can be evaluated as ​A=P+Prt
A=P(1+rt)
P=A/1+rt


The value of P found after evaluating is  A/1+rt.

How To Solve Exercise 1.5 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 17 Essential Question Answer

To solve absolute value equations algebraically, first, isolate the absolute value expression on one side of the equation the same way you would isolate a variable.

Then use the rule: If x=a (where a is a positive number), then x=a OR x=−a.

To solve an absolute value equation, isolate the absolute value on one side of the equation. Then set its contents equal to both the positive and negative value of the number on the other side of the equation and solve both equations.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 17 Exercise 1 Answer

Given: The expression is ​|x+2|=3.

To find The value of ​x+2 that makes the equation true and d write it in an absolute form. Evaluate to get the answer.

We know that an absolute value equation can have either a positive or a negative value:

Set up the equations:

|x+2|=3:(x+2)=3…………….(1)​
|x+2|=3:(x+2)=−3…………….(2) ​

The required solution is (x+2)=3 and(x+2)=−3.

Given: The expression is x+2=3.

To find The solution of the absolute value equation. Evaluate to get the answer.

Solve for x using Eq(1)and Eq(2):

​1. |x+2|=3:​(x+2)=3
x=3−2
x=1

2. ​​|x+2|=3:(x+2)=−3
x=−3−2
x=−5

The required solutions are x=1 and x=−5.

We can use linear equations to solve absolute value problems by separating the equations into positive and negative components.

We can use linear equations to solve absolute value problems by separating the equations into positive and negative components.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4

Big Ideas Math Algebra 1 Chapter 1 Exercise 1.4 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 18 Exercise 3 Answer

 

Given: The absolute expression is x+2=3.

To find The values of absolute equations using a spreadsheet. Evaluate to get the answer.

We can solve the values for x+2 using a spreadsheet as follows:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations table 2

The required solution is :

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations table 1

Given: The expression is x+2=3.

To compare: The solutions of the spreadsheet with the solutions in Explorations 1 and 2.

We observe that the value for(x+3)=3 and =-3 will be as follows:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations table 3

The required answer is :

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations table 4

The absolute value equation is x+2=3.

We can solve absolute value equations using a spreadsheet by plugging in the values of x in the equation.

We can solve absolute value equations using a spreadsheet by plugging in the values of in the equation.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 18 Exercise 4 Answer

We can solve an absolute value equation using these steps:

Separate the absolute value equation to positive and negative components.

Set up these equations.

Solve for the values of the positive and negative components.

We separate the absolute value equation into positive and negative components and solve for these values.

Solving Linear Equations Exercise 1.4 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 18 Exercise 5 Answer

We are given the absolute value equation: |f(x)|=c.

Solving the equation algebraically means solving two equations:​ f(x)=c and (x)=c. It is easy that we do not need graphics or tables, but we have to solve two equations.

Solving the equation graphically means drawing the number line and locating the point for which f(x)=0.
Then we locate the points c units from the point we determined. It is easy because we only solve one equation, but we have to draw a graph.

Solving the equation numerically means building a table of values and trying to find those values of x for which|f(x)|=c. It is easy because we don’t solve any equation and we don’t draw a graph, but we have to build a table and sometimes search among a big number of values.

On solving the equation algebraically we have to solve two equations. On solving equations graphically we only solve one equation, but we have to draw a graph. To solve the equation numerically we have to build a table and sometimes search among a big number of values.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 21 Exercise 4 Answer

Given: The expression is​ |d/3|=3.

To plot: The graph of solution on a number line. Evaluate to get the answer.

On solving the given equation :

solve for d: |d/3|=3

1. |d/3|=3:  (d/3)=3
d=3.3
d=9

2. |d/3|=3:  (d/3)=−3
d=−3⋅3
d=−9

Plot d=9 and d=−9

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 1

The required solution is d=9 and d=−9.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 2

Algebra 1 Student Journal Chapter 1 Exercise 1.4 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 21 Exercise 5 Answer

Given: The expression is  3 |2x+5|+10=37.

To plot: The graph of solution on a number line. Evaluate to get the answer.

On solving the given equation :

solve for x,

​3|2x+5|+10=37
3(2x+5)=37−10       (distribute)
3.2x+3.5&=27         (simplify)
6x+15=27               (isolate x)
6x=27−15
6x=12
x=12/6
x=2

Solve for x:

​3(2x+5)=−(37−10)          (distribute)
3⋅2x+3⋅5=−2                  (simplify )
6x+15=−27                    (isolate x)
6x=−27−15
6x=−42
x=−42/6
x=−7

Plot x=2 and x=−7 on a number line :

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 3

Final answer The required solution is and.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute value equations image 3

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 21 Exercise 6 Answer

Given: The expression is |20x|=|4x+16|.

To plot: The graph of solution on a number line. Evaluate to get the answer.

On solving the given equation :

solve for x:

​|20x|=|4x+16|
20x=4x+16
20x−4x=16
16x=16
16x=16
x=16/16
x=1

Solve for x:

​20x =−(4x+16)​ (distribute)
20x=−4x−16​ (isolate x)
20x+4x=−16
24x=−16
24x=−16
x=−16/24
x=−2/3

Plot x=1 and x=−2/3:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 5

The required solutions are x=1 and x=−2/3.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 6

Big Ideas Math Linear Equations Exercise 1.4 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 21 Exercise 7 Answer

Given: The expression is |p+4|=|p−2|.

To plot: The graph of solution on the number line. Evaluate to get the answer.

On solving the equation :

Solve for p :

|p+4|=|p−2|
(p+4)=±(p−2)
(p+4)=(p−2)​           (isolate p)
​p−p=−2−4             (​simplify)
0=−6
It has no solution.

Solve for p:

(p+4)=−(p−2)       (distribute)
p+4=−p+2            (isolate p)
p+p=2−4
2p=−2
p=−2/2
p=−1

Plot p=−1:

The required solution is p=−1.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 8

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 21 Exercise 8 Answer

Given: The expression is​|4q+9|=|2q−1|.

To plot: The graph of solution on the number line. Evaluate to get the answer.

On solving the given equation :

solve for q:

​|4q+9|=|2q−1|
(4q+9)=±(2q−1)
(4q+9)=(2q−1)​               (isolate q)
4q−2q=−1−9​                 (simplify )
2q=−10
q=−10/2
q=−5

solve for q:

(4q+9)=−(2q−1)        (distribute)
(4q+9)=−2q+1          (isolate q)
4q+2q=1−9             (simplify)
6q=−8
q=−8/6
q=−4/3

Plot q=−5 and q=−4/3:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 9

The required solutions are q=−5 and q=−4/3.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 10

Chapter 1 Exercise 1.4 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.4 Page 21 Exercise 9 Answer

Given: The expression is |2x−7|=|2x+9|.

To plot: The graph of solution on the number line. Evaluate to get the answer.

On solving the equation :

solve for x,

​|2x−7|=|2x+9|
(2x−7)=±(2x+9)
​2x−7=2x+9                (isolate x)
2x−2x=9+7​               (simplify )
0=16
It has no solution.

Solve for x:

2x−7=−(2x+9)          (distribute)
2x−7=−2x−9            (isolate)
2x+2x=−9+7           (simplify)
4x=−2
x=−2/4
x=−1/2

Plot x=−1/2:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .4 Solving Absolute valule equations image 11

The required solution is x=−1/2.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 12 Exercise 1 Answer

Given: The perimeters of 2 Figures are equal.
To find The perimeter of each polygon and the value of x. Evaluate to get the answer.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .3 Solving Equations with variables on both sides image 1

Using the concept of equal perimeters, it can be written as
5+x+5+2+x+2=3/2x+5+4+(3/2x−4)+3+5−3
2x+14=3x+10.
2x−2x+14=3x−2x+10     (Subtract 2x on both sides)
14=x+10
4=x.

The perimeter is 5+4+5+2+4+2=22m.

The obtained value is x=4 and the perimeter is 22m.

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 12  Exercise 2  Answer

Given:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .3 Solving Equations with variables on both sides image 2

 

To find Value of x. Evaluate to get the answer.

Using the fact that area and perimeter are equal,

⇒ ​5+5+2(4)+x=3x/2+4x
⇒ 18+x=3x/2+4x
⇒ 36+2x=3x+8x
⇒ 36+2x=11x
⇒ 36=11x−2x
⇒ 36=9x

∴x=4

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3

The area and perimeter obtained is 5+5+2(4)+4=22.

After evaluating the area obtained is 22.

Given:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .3 Solving Equations with variables on both sides image 3

To find Value of x. Evaluate to get the answer.

Using the concept of area and perimeter are equal,

​⇒ 2x+6+4(1)=6x−2[2(1)]
⇒ 2x+10=6x−4
⇒ 2x+10−2x=6x−4−2x
⇒ 10=4x−4
⇒ 4x=14
⇒ x=14/4

∴ x=3.5

The area is 2(3.5)+6+4(1)=17.

After evaluating the area and perimeter obtained is 17.

Given:

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .3 Solving Equations with variables on both sides image 4

To find The area and perimeter of the. Evaluate to get the answer.

Using the given, we solve for x.

​2(2)+2x+π⋅2=[2(2)]x+π((22​)/ 2 )
4+2x+2π=4x+2π
4+2x=4x
2x=4
x=2

The area obtained is

2(2)+2(2)+π.2=8+2π

Perimeter =2.(l+b)+π.r2 where r=2
=2(2+4)+3.14.2.2
=12+12.56
=24.56

The area obtained is 8+2π

The perimeter obtained is 24.56

Big Ideas Math Algebra 1 Chapter 1 Exercise 1.3 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 13 Exercise 4 Answer

Given: The equations should be different.

To find: The 3 equations having variables on both sides. Evaluate to get the answer.

The 3 equations can be written as:
​(1) ​3x−7=4x−9
(2) 11−2x=7x+3
(3) 5−6x=9x−4

Solving for the equations,

  1. 3x−7=4x−9
    3x−4x=−9+7
    −x=−2
    x=2
  2. 11−2x=7x+3
    −2x−7x=3−11
    −9x=−8
    −9x/−9=−8/−9
    x=8/9
  3.  5−6x=9x−4
    −6x−9x=−4−5
    −14x=−9
    −14x/-14=−9/−14
    x=9/14

The obtained values for the equations are ​x​=2, x=8/9, x=9/14

Big Ideas Math Algebra 1 Chapter 1 Exercise 1.3 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 15 Exercise 3 Answer

Given: The expression is​ 3k+45 =8k+25.

To find The value of constant k. Evaluate to get the answer.

On solving the given equation :

Solve for k :

⇒ ​3k+45=8k+25
⇒  3k−8k=25−45
⇒  −5k=−20
⇒  −5k/−5=−20/−5
∴ k=4

The required solution is k=4.

Algebra 1 Student Journal Chapter 1 Exercise 1.3 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 15 Exercise 4 Answer

Given: The expression is 3/4(48−16x)=4(4+2x).

To find The value of constant x. Evaluate to get the answer.

On solving the given equation :

Solve for x :

3/4(48−16x)=4(4+2x)​                  (distribute)

3/4.48−3/4⋅(−16x)=4⋅4+4⋅2x​       (simplify)

36−12x=16+8x

12x−8x=16−36​                             (isolate  x)

x−20x=−20

−20x/−20=−20/−20

x=1

The required solution is x=1.

Algebra 1 Student Journal Chapter 1 Exercise 1.3 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 15 Exercise 5 Answer

Given: The expression is​ 5q+6=2q−2+q
To find The constant value q. Evaluate to get the answer.

On solving the given equation :

Solve for q :

⇒ ​5q+6=2q−2+q
⇒ 5q−2q−q=−2−6
⇒ 2q=−8
⇒ q=-8/2
∴ q=−4

The required solution is q =−4.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 7 Answer

Given: The expression is​ 6a−4 =3a+5.
To find The constant value of a. Evaluate to get the answer.

On solving the given equation :

Solve for a :

⇒ ​6a−4=3a+5
⇒ 6a−3a=5+4
⇒ 3a=9
⇒ a=9/3

∴a=3

The required solution is a=3.

Big Ideas Math Linear Equations Exercise 1.3 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 8 Answer

Given: The expression is​  2(4b−6)=4(3b−7).

To find: The constant value of evaluate to get the answer.

On solving the given equation :

Solve for b:

⇒ ​2(4b−6)=4(3b−7)​               (distribute)
⇒  2.4b−2.6=4.3b−4.7            (simplify)
⇒  8b−12=12b−28           ​     (isolate b)
⇒  8b−12b=−28+12
⇒  −4b=−16
⇒  b=−16/−4

∴ b=4

The required solution is b=4.

Chapter 1 Exercise 1.3 Step-By-Step Guide Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 9 Answer

Given : The given expression​ 8(2r−3)−r =3(3r+2)

To find The constant value of r.Evaluate to get the answer.

On solving  the  given equation:

Solve for r :

​⇒  8(2r−3)−r =3(3r+2)​              (distribute)
⇒  8.2r−8.(−3)=3.3r+3.2​            (simplify)
⇒  16r−24−r=9r+6​                   (isolate r)
⇒  16r−r−9r=6+24
⇒  6r=30
⇒  r=30/6

∴r=5

The required solution is r=5.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 10 Answer

Given : The expression is​ 3x−8(2x+3) =−6(2x+5)

To find: The constant value of x Evaluate to get the answer.

On solving the given equation :

Solve for x :

⇒ ​3x−8(2x+3) =−6(2x+5)                  (distribute)
⇒  3x−(8.2x+8.3)=−6.2x+(−6.5)          (simplify)
⇒  3x−(16x+24)=−12x−30
⇒  3x−16x−24=−12x−30​                     (isolate x)
⇒  3x−16x+12x=−30+24
⇒  3x-4x=−6

∴ −x=6

The required solution is x=6.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 11 Answer

Given : The expression is ​ 6(4s+12) =8(3s−14)

To find: The constant value of s Evaluate to get the answer.

On solving the given equation :

Solve for s :

​6(4s+12) =8(3s−14)​        (distribute)
6⋅4s+6⋅12=8⋅3s−8⋅14​      (simplify)
24s+72=24s−112​             (isolate s)
24s−24s=−112−72
0=−184
s has no solution.

This equation is not possible to exist. s has no solution.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 12 Answer

Given: The expression is 16f+24=8(2f+3).

To find The number of solutions possible for the equation. Evaluate to get the answer.

On solving the given equation :

Solve f:

​16f+24=8(2f+3)​          (distribute)
16f+24=16f+24

We observe that both sides of the equation are equal, hence, if we use any value for f, then we can have many solutions for this equation.

We can have many values for f, then the equation has infinitely many solutions.

Exercise 1.3 Solutions Big Ideas Math Algebra 1

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 13 Answer

Given : The expression is 1/2(10+12n)=1/3(15n+15)

To find The number of solutions possible. Evaluate to get the answer.

On solving the given equation :

Solve for n :

1/2(10+12n)=1/3(15n+15)             (distribute)
​1/2.10+1/2.12n=1/3.15n+1/3.15   (simplify)
5+6n=5n+5                                  ​ (isolate n)
6n−5n=5−5
n=0

Only one solution is possible since n=0, this is a linear equation.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 14 Answer

Given : The expression is 2/3(6j+9)=3j+7

To find: The number of solutions is possible for the equation. Evaluate to get the answer.

On solving the given equation :

Solve j :

2/3(6j+9)=3j+7               (distribute)
2/3.6j+2/3.9=3j+7           (simplify )
4j+6=3j+7                       (isolate j)
4j−3j=7−6
j=1

Only one solution is possible since j=1.

Big Ideas Math Chapter 1 Exercise 1.3 Walkthrough

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.3 Page 16 Exercise 15 Answer

Given :

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 1 .3 Solving Equations with variables on both sides image 5

The value of the surface area of a rectangular prism is equal to the value of the volume of the rectangular prism

To find The value of x. Evaluate to get the answer.

The surface area of a rectangular prism is given by:

⇒ S=2(lw+wh+lh)
Substitute l=6,w=x, and h=6 :
⇒ ​S=2(6⋅x+x⋅6+6⋅6)
⇒ S=2(6x+6x+36)
⇒ S=2(12x+36)
⇒ S=24x+72

The volume of a rectangular prism is given by:

V=Bh

where B is the area of the base and h is the height. From, the base is a rectangle so B=6x. We also have h=6:

​⇒ V=(6x)(6)
⇒ V=36x

The values of the surface area and volume are equal so :

⇒ S=V
⇒ 24x+72=36x
⇒ 72= 36x−24x
⇒ 72=12x
⇒ 12x= 72
⇒ x=72/12

∴ x=6 units

The required value of x is x=6 units.

How To Solve Exercise 1.3 Big Ideas Math