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.
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:
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
The area and perimeter obtained is 5+5+2(4)+4=22.
After evaluating the area obtained is 22.
Given:
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:
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,
- 3x−7=4x−9
3x−4x=−9+7
−x=−2
x=2 - 11−2x=7x+3
−2x−7x=3−11
−9x=−8
−9x/−9=−8/−9
x=8/9 - 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 :
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