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Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 1.1
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Exercise 45 Page 27

Practice makes perfect
a To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side by using the Properties of Equality.

75=14y+5
SubEqn

LHS-5=RHS-5

70=14y
RearrangeEqn

Rearrange equation

14y=70
DivEqn

.LHS /14.=.RHS /14.

y=5

The solution to the equation is y=5. We can check our solution by substituting it into the original equation.

75 = 14y+5
Substitute

y= 5

75 ? = 14( 5)+5
â–¼
Simplify
Multiply

Multiply

75 ? = 70+5
AddTerms

Add terms

75=75 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

b To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side by using the Properties of Equality.

- 7r+13=- 71
SubEqn

LHS-13=RHS-13

- 7r = - 84
DivEqn

.LHS /(- 7).=.RHS /(- 7).

- 7r/- 7=- 84/- 7
DivNegNeg

- a/- b=a/b

7r/7=84/7
CalcQuot

Calculate quotient

r=12

The solution to the equation is r=12. We can check our solution by substituting it into the original equation.

- 7r+13=- 71
Substitute

r= 12

- 7( 12)+13 ? = - 71
â–¼
Simplify
MultNegPos

(- a)b = - ab

- 84 +13 ? = - 71
AddTerms

Add terms

- 71 = - 71 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

c To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side by using the Properties of Equality.

3a+11=7a-13
AddEqn

LHS+13=RHS+13

3a+24=7a
SubEqn

LHS-3a=RHS-3a

24=4a
RearrangeEqn

Rearrange equation

4a=24
DivEqn

.LHS /4.=.RHS /4.

a=6

The solution to the equation is a=6. We can check our solution by substituting it into the original equation.

3a+11=7a-13
Substitute

a= 6

3( 6)+11 ? = 7( 6)-13
â–¼
Simplify
Multiply

Multiply

18+11 ? = 42-13
AddSubTerms

Add and subtract terms

29=29 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

d To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side by using the Properties of Equality.

2m+m-8=7
AddTerms

Add terms

3m-8=7
AddEqn

LHS+8=RHS+8

3m=15
DivEqn

.LHS /3.=.RHS /3.

m=5

The solution to the equation is m=5. We can check our solution by substituting it into the original equation.

2m+m-8=7
Substitute

m= 5

2( 5)+ 5-8 ? = 7
â–¼
Simplify
Multiply

Multiply

10+5-8 ? = 7
AddSubTerms

Add and subtract terms

7=7 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

Subchapter links expand_more
arrow_right 1.1.1 Creating a Quilt Using Symmetry p.7
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Creating a Quilt Using Symmetry 5 (Page 7)
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arrow_right 1.1.2 Making Predictions and Investigating Results p.11-15
Making Predictions and Investigating Results 14 (Page 11)
Making Predictions and Investigating Results 15 (Page 12)
Making Predictions and Investigating Results 16 (Page 13)
Making Predictions and Investigating Results 17 (Page 14)
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arrow_right 1.1.3 Perimeter and Area of Enlarging Tile Patterns p.16-17
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arrow_right 1.1.4 Logical Arguments p.21-22
32
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34
35
36
arrow_right 1.1.5 Building a Kaleidoscope p.26-27
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Building a Kaleidoscope 45 (Page 27)
46