Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 5 Solving Systems Of Linear Equations Exercise 5.6

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 5 Solving Systems Of Linear Equations

 

Page 157   Essential Question   Answer

In the above-given question, It is asked how one can graph a linear inequality in two variables.
Since a solution of a linear inequality in two variables is an ordered pair(x,y) for which the inequality is true.

The graph of a linear inequality in two variables shows all the solutions of the inequality in a coordinate plane.
We will give an example of inequality to prove the above.

Consider the inequality in two variables y ≥ −x − 5.
The graph of the inequality y ≥ −x − 5 is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 1

 

Since y ≥ −x−5.

Put x = −2 and y = −2 in the above inequality, and we get

−2≥−(−2)−5

⇒ −2≥2−5

⇒ −2≥−3

Hence the inequality y ≥−x−5 is true for (−2,−2).
Therefore (−2,−2)  is the solution for the inequality y ≥−x−5.

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 5 Solving Systems Of Linear Equations Exercise 5.6

 

1. Graph of the inequality y ≥ −x−5 

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 2

 

2. (-2,-2) is the solution of the inequality y ≥ −x−5.

 

Page 157  Exercise 1  Answer

In the above-given question, we were asked to write an equation represented by the dashed line in the given graph.

To solve the above we will use the Point slope form to write an equation of a line.
We calculate the slope of the dashed line given in the graph and consider one of the points on the line.

From the given graph, we have the points (3,0) and (0,−3) on the dashed line.
Therefore the slope of the dashed line is given by

m=\(\frac{-3-0}{0-3} \)

\(\frac{-3}{-3}\)

 

=1

Therefore the slope(m) of the dashed line is 1.

 

Since the slope of the dashed line is 1 and (0,−3) is one of the points on the dashed line.
Therefore the equation of the dashed line by the point-slope form is given by
y−y1= m(x−x1) ……(1)

Put  x1 = 0, y1 = −3 and m = 1 in the above equation, we get

y−(−3) = 1(x−0)

⇒ y+3 = x

⇒ y = x−3

Therefore y = x−3 is the equation of the dashed line.

An equation represented by the dashed line is y = x−3.

 

In the above-given question, the solutions of the inequality are represented by the shaded region below the dashed line y = x−3.

We were asked, to describe the solutions to the inequality represented by the shaded region.
We make use of the given graph to solve the above.

Since the line y = x−3 is represented by the dashed line.
Therefore the shaded region includes all points below the line y = x−3 but does not contain the points on the line y = x−3.

Hence the solutions of the inequality include points below the line y = x−3  and does not contain the points on the line y = x−3.


The solutions of an inequality represented by the shaded region include the points below the dashed line y=x−3 that does not contain the points on the line.

 

In the above-given question, we were asked to write an inequality that represents the above graph.
We make use of the given graph to solve this.

In the above graph, the solutions of the inequality are represented by the shaded region that is below the dashed line  y=x−3.

Since the solutions of the inequality include all the points below the dashed line and no points on the dashed line.

Hence the values of y are always less than  x−3.
Therefore we use strictly less than (<)  sign to represent the graph.
Therefore corresponding inequality that represents the graph is y<x−3.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 3

 

The inequality represented by the graph is y< x−3.

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 4

 

Page 157  Exercise 2  Answer

Given: Inequality y ≥ \(\frac{1}{4} x-3\).

We asked to enter the equation y\(=\frac{1}{4} x-3\) inside the graphing calculator.
To solve the above we simply enter the equation y=\(\frac{1}{4} x-3\) inside the calculator to get the desired result.

By entering the equation y=\(\frac{1}{4} x-3\) into the calculator, we get the graph of y=\(\frac{1}{4} x-3\)
Therefore the graph of y=\(\frac{1}{4} x-3\)after entering the corresponding equation inside the calculator is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 5

 

Graph of the equation  y\(=\frac{1}{4} x-3\) after entering the equation y\(=\frac{1}{4} x-3\) inside the calculator is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 6

 

Given:  The Inequality  y \(\geq \frac{1}{4} x-3\)

We asked to graph y \(\geq \frac{1}{4} x-3\).
Using a graphing calculator also to check the same by considering a point inside the region.
To solve the above we enter the equation y \(=\frac{1}{4} x-3\) inside the calculator then accordingly shade the region.

 

Enter the equation y=\(\frac{1}{4} x-3\) inside the graphing calculator. After entering the above equation we get the graph of y=\(\frac{1}{4} x-3\) is given by

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 7

 

Since inequality y \(\geq \frac{1}{4} x-3\) has the symbol≥.
Therefore the shaded region is above the graph of y=\(\frac{1}{4} x-3\)
Therefore the graph of y \(\geq \frac{1}{4} x-3\) is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 8

 

Since  y\(\geq \frac{1}{4} x-3\)

Put(0,0) in the above inequality, and we get

⇒  \(\geq \frac{1}{4}(0)-3\)

⇒  0 ≥ 0−3

⇒  0 ≥−3

Hence the point(0,0) lies in the region y \(\geq \frac{1}{4}(0)-3\)
Therefore the point(0,0) is the solution of the inequality y\(\geq \frac{1}{4}(0)-3\)

 

1. Graph of y\(\geq \frac{1}{4}(0)-3\) by using a graphing calculator is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 9

2. The inequality y\(\geq \frac{1}{4}(0)-3\) is true for(0,0).

 

Page 158  Exercise 3  Answer

Given: The Inequality y > x + 5.

We asked to graph the above inequality, and also to check the same using graphing calculator.
To solve the above we graph the equation y = x + 5 then accordingly shade the region.

 

Since the inequality y > x + 5 does not contain the equal to(=) sign.
Therefore we graph the equation y = x + 5 with a dashed line.
Therefore The graph of the equation y = x + 5 is given by

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 10

 

Since the inequality y>x+5 contains the strictly less then (>) sign.
Therefore the shaded region will be above the line y =x + 5 that does not contain the line.
Therefore the graph of the inequality  y > x + 5 is given by

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 11

 

Since y >  x + 5.

Put x = −1 and y = 1 in the above inequality, and we get

⇒ 1 >−1 + 5

⇒ 1 > 4

Hence(−1,1) does not satisfy the inequality y > x + 5.
Therefore(−1,1) is not a solution to the inequality y > x + 5.

 

1. Graph of y> x+5 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 12

 

2.Graph of y>x+5 using a graphing calculator is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 13

 

3. (-1,1) is not a  solution  to the Inequality  y > x+5

Given: The  Inequality y≤\(\frac{-1}{2} x+1\)

We asked to graph the above inequality, and also to check the same using graphing calculator.
To solve the above we graph the equation y=\(\frac{-1}{2} x+1\) then accordingly shade the region.

We graph the linear equation y=\(\frac{-1}{2} x+1\)
with the continuous line, since the inequality contains the equal to(=) sign.
Therefore the graph of y=\(\frac{-1}{2} x+1\) is given by

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 14

 

Since the inequality y≤\(\frac{-1}{2} x+1\) contains the sign≤.
Therefore the shaded region will be below the line y=\(\frac{-1}{2} x+1\)
Therefore the graph of y ≤\(\frac{-1}{2} x+1\) is given by

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 15

 

Since y≤ \(\frac{-1}{2} x+1\)

Put x = −1 and y = −1 in the above inequality, and we get

−1≤ \(\frac{-1}{2}(-1)+1\)

⇒ −1≤ \(\frac{-1}{2}+1\)

⇒ −1≤\(\leq \frac{3}{2} \approx 1.5 \)

Hence the point(−1,−1) satisfy the inequality y≤\(\frac{-1}{2} x+1\)
Therefore(−1,−1) is the solution for the inequality y≤\(\frac{-1}{2} x+1\).

 

1. Graph of y=\(\frac{-1}{2} x+1\) is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 16

 

2. Graph of y≤\(\frac{-1}{2} x+1\) using a graphing calculator is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 17

3.The inequality y≤\(\frac{-1}{2} x+1\) is true for(−1,−1).

 Given: The inequality is y ≥−x−5

We have to graph this inequality.

For that, we have to draw a line y = −x−5 and shade the solution region which is above the line.
We can check this solution by using a random point.

We have y ≥−x−5
To graph this inequality
First, we have to draw a line y=−x−5 and shade the solution region which is above the line. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 18

 

To check this solution Let us take a random point (0,0) So here,​ x = 0 y = 0 Therefore

​⇒  y ≥−x−5

​⇒  0≥−0−5

​⇒  0≥−5

Here,(0,0) is in the solution therefore, the solution is true.

 

The graph of the inequality y ≥−x−5 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 19

 

Page 158   Exercise 5  Answer

Let x is the number of books and y is the number of pencils.
A person has  $15 and wants to buy notebooks and pencils.

If the price of a notebook is  $2 and the price of a pencil is  $1.
Then the items of each type the person can buy are  2x + y ≤ 15.
An example of a real-life situation that can be modeled using a linear inequality in two variables is.

A person has $15 and wants to buy notebooks and pencils.
If the price of a notebook is $2 and the price of a pencil is $1.
Then the items of each type the person can buy are  2x + y ≤ 15.

 

Page 160  Exercise 2  Answer

Given: The given inequality is x−y ≥ 2  and the ordered pair is (5,3).

We have to find whether the ordered pair is a solution to the inequality.
For that, we have to check whether the given ordered pair satisfies the inequality.

We have the inequality x−y ≥ 2. The given ordered pair is(5,3)i.e.

⇒  ​​x−y​ ≥ 2

⇒ 5−3 ≥ 2

⇒ 2 ≥ 2

Here this solution is true.

The given ordered pair(5,3) is a solution to the inequality  x−y ≥ 2

 

Page 160   Exercise 3  Answer

 Given:  The inequality is​ x + 2y ≤ 4 and the ordered pair is (−1,2).

We have to find whether the ordered pair is a solution to the inequality.
For that, we have to check whether the given ordered pair satisfies the inequality.

We have the inequality​ x + 2y ≤ 4. The given ordered pair is (−1,2) i.e.

⇒ ​−1 + 2 × 2 ≤ 4

⇒ −1 + 4 ≤ 4

⇒ 3 ≤ 4

Here this solution is true.

The given ordered pair(−1,2) is a solution to the inequality x + 2y ≤ 4.

 

Page 160  Exercise 4  Answer

 Given:  The inequality is​ 5x + y < 7 and the ordered pair is (2,−2).

We have to find whether the ordered pair is a solution to the inequality.
For that, we have to check whether the given ordered pair satisfies the inequality.

We have the inequality​ 5x + y < 7. The given ordered pair is (2,−2) i.e.​​

⇒  5×2+(−2) < 7

⇒  10−2 < 7

⇒  8 < 7

Here this solution is not true.

The given ordered pair(2,−2) is not a solution to the inequality 5x + y < 7.

 

Page 160  Exercise 6  Answer

 Given:  The inequality is​  −x−2y ≥ 5 and the ordered pair is (−2,−3).

We have to find whether the ordered pair is a solution to the inequality.
For that, we have to check whether the given ordered pair satisfies the inequality.

We have the inequalit  y​−x−2y ≥ 5 .The given ordered pair is (−2,−3)i.e.​​

⇒  −x−2y ≥ 5

⇒  −(−2)−2(−3) ≥ 5

⇒  2−(−6) ≥ 5

⇒  2 + 6 ≥ 5

⇒  8 ≥ 5

​Here this solution is true.

The given ordered pair(−2,−3) is a solution to the inequality y−x−2y≥5.

 

Page 160  Exercise  7 Answer

 Given:  The inequality is y < 4.

We have to graph this inequality.
For that, we have to draw a line y = 4  and shade the solution region which is below the line.

We have y<4
To graph this inequality
First, we have to draw a line y=4 and shade the solution region which is below the line. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 20

 

The graph of the inequality y<4 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 21

 

Page 160  Exercise 8  Answer

 Given: The inequality is y >−1.

We have to graph this inequality.
For that, we have to draw a line y = −1 and shade the solution region which is above the line.

We have y >−1
To graph this inequality
First, we have to draw a line y >−1
and shade the solution region which is above the line.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 22

 

The graph of the inequality y >−1

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 23

 

Page 160  Exercise 9  Answer

 Given: The inequality is x >3.

We have to graph this inequality.
For that, we have to draw a line x=3 and shade the solution region which is greater than x=3.

We have x >3
To graph this inequality
First, we have to draw a line x = 3 and shade the solution region greater than x = 3. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 24

The graph of the inequality x >3 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 25

 

Page 160  Exercise 10  Answer

 Given: The inequality is x ≤−1.

We have to graph this inequality.
For that, we have to draw a line x =−1 and shade the solution region which is less than or equal to−1.

We have x ≤−1
To graph this inequality
First, we have to draw a line x =−1 and shade the solution region less than or equal to−1. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 26

 

The graph of the inequality x = −1 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 27

 

Page 160  Exercise 11  Answer

 Given: The inequality is y <−2.

We have to graph this inequality.
For that, we have to draw a line y = −2 and shade the solution region which is below the line.

We have y <−2
To graph this inequality
First, we have to draw a line y <−2 and shade the solution region which is below the line. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 28

 

The graph of the inequality y <−2 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 29

 

Page 160  Exercise 12  Answer

 Given: The inequality is x >−2.

We have to graph this inequality.
For that, we have to draw a line x = −2 and shade the solution region which is greater than−2.

We have x >−2
To graph this inequality
First, we have to draw a line x =−2 and shade the solution region greater than−2. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 30

 

The graph of the inequality x >−2 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 31

 

Page 161  Exercise 13  Answer

 Given: The inequality is y< 3x + 1.

We have to graph this inequality.
For that, we have to draw a line y = 3x + 1 and shade the solution region which is below the line.

We have y < 3x + 1
To graph this inequality
First, we have to draw a line y = 3x + 1 and shade the solution region which is below the line. i.e.

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 32

 

The graph of the inequality y = 3x + 1 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 33

 

Page 161  Exercise 15  Answer

Given: x−y < 2

To find graph the inequality in a coordinate plane.
First, get y alone on one side, then consider the related equation obtained by changing the inequality.

Sign to an equality sign, then the graph of this equation is a line.
If the inequality is strict graph a dashed line and if the inequality is not strict graph a solid line.

The given inequality can be rewritten as

⇒  ​x−y < 2

⇒ ​ −y < 2 − x

⇒ ​ y > −2 + x

The related equation is y = −2 + x.

 

Since the inequality sign is strict, so draw a dashed line for the related equation.
Since the inequality sign is greater than, so shade the region above the boundary line.
So, the graph of the inequality x−y < 2 is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 34

 

The graph of the inequality x−y < 2 in a coordinate plane is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 35

 

 

Page 161  Exercise 16  Answer

Given: x + y ≥−3

To find  graph the inequality in a coordinate plane.
First, get y alone on one side, then consider the related equation obtained by changing the inequality.

Sign to an equality sign, then the graph of this equation is a line.
If the inequality is strict graph a dashed line and if the inequality is not strict graph a solid line.

The given inequality can be rewritten as

​⇒ ​ x + y ≥ −3

⇒ ​ y ≥ −3−x

The related equation is  y = −3−x.

 

Since the inequality sign is not strict, so draw a solid line for the related equation.
Since the inequality sign is greater than or equal to, so shade the region above the boundary line.
So, the graph of the inequality  x + y ≥−3 is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 36

 

The graph of the inequality x + y ≥ −3 in a coordinate plane is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables garph 37

 

Page 161  Exercise 17 Answer

Given: x + 2y < 4

To find a graph of the inequality in a coordinate plane.
First, get y alone on one side, then consider the related equation obtained by changing the inequality.

Sign to an equality sign, then the graph of this equation is a line.
If the inequality is strict graph a dashed line and if the inequality is not strict graph a solid line.

The given inequality can be rewritten as

⇒ ​ ​x + 2y < 4

⇒ ​ 2y < 4−x

⇒ ​ y<\(2-\frac{x}{2}\)

​The related equation is y=\(2-\frac{x}{2}\)

 

Since the inequality sign is strict, so draw a dashed line for the related equation.
Since the inequality sign is less than, so shade the region below the boundary line.
So, the graph of the inequality  x + 2y < 4 is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables graph 38

 

The graph of the inequality x + 2y < 4 in a coordinate plane is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables graph 39

 

Page 161  Exercise 18  Answer

Given: The linear function, −2x+3y>6.

Let’s assume it a linear equation, and Find the coordinates of the x, and y-axis intersecting points.
Plot them on the graph and draw a dotted line connecting those points.
Shade the area above that linear equation.

Let, the linear equation be −2x + 3y = 6.
Point of intersecting X – axis is (x,0)

⇒−2(x)+3(0)=6

⇒−2x=6

⇒x=−3

∴  The point of the intersecting X-axis is (-3,0)

 

Plot the points in graph

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables graph 40

 

The graph of given inequality is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables graph 41

 

Page 161  Exercise 19  Answer

Given:  Profit on sale of each digital camera x is 100 and cell phone y is 50.

By selling digital cameras and cell phones store wants to make profit at least 300.
Find a linear equation and plot the linear equation on a graph.
Find any two points justifying the equation and interpret

Let, the number of digital cameras sold be x, cellphones sold be y.
Then, the linear inequality will be  100x + 50y ≥ 300.

Plot the inequality on graph
Identify any two points covered in the shaded area

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables graph 42

 

At (1,4), i.e. one unit of digital camera and 4 units of cell phones sold profit will be

=100(1) + 50(4)

=300

 

At  (4,2),  i.e. 4 units of Digital cameras and 2 units of cell phones sold profit will be

=100(4) + 50(2)

=500(>300)

 

A linear inequality is 100x +50y ≥ 300. Graph of the above linear inequation is

 

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 5.6 Graphing Linear Inequalities In Two variables graph 43

 

Any point covered under the shaded area will solve the given inequation

 

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