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Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 1.2
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Exercise 79 Page 43

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.
4p+8p+12=- 48
AddTerms

Add terms

12p+12=- 48
SubEqn

LHS-12=RHS-12

12p=- 60
DivEqn

.LHS /12.=.RHS /12.

p=- 5
The solution to the equation is p= - 5. We can check our solution by substituting it into the original equation.
4p+8p+12=- 48
Substitute

p= - 5

4( - 5) +8( - 5) +12 ? = - 48
â–Ľ
Simplify
MultPosNeg

a(- b)=- a * b

- 20 -40 +12 ? = - 48
SubTerm

Subtract term

- 60 +12 ? = - 48
AddTerms

Add terms

- 48=- 48 âś“
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.
9w+33=12w-27
AddEqn

LHS+27=RHS+27

9w+60=12w
SubEqn

LHS-9w=RHS-9w

60=3w
RearrangeEqn

Rearrange equation

3w=60
DivEqn

.LHS /3.=.RHS /3.

w=20
The solution to the equation is w=20. We can check our solution by substituting it into the original equation.
9w+33=12w-27
Substitute

w= 20

9( 20)+33 ? = 12( 20) - 27
â–Ľ
Simplify
Multiply

Multiply

180+33 ? = 240-27
AddSubTerms

Add and subtract terms

213 = 213 âś“
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.
- 2x+7=- 21
SubEqn

LHS-7=RHS-7

- 2x=- 28
DivEqn

.LHS /(- 2).=.RHS /(- 2).

- 2x/- 2=- 28/- 2
DivNegNeg

- a/- b=a/b

2x/2=28/2
CalcQuot

Calculate quotient

x=14
The solution to the equation is x=14. We can check our solution by substituting it into the original equation.
- 2x+7=- 21
Substitute

x= 14

- 2( 14)+7 ? = - 21
â–Ľ
Simplify
MultNegPos

(- a)b = - ab

- 28+7 ? = - 21
AddTerms

Add terms

- 21 = - 21 âś“
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.
25=10y+8
SubEqn

LHS-8=RHS-8

17=10y
RearrangeEqn

Rearrange equation

10y=17
DivEqn

.LHS /10.=.RHS /10.

10y/10 = 17/10
CalcQuot

Calculate quotient

y=17/10
The solution to the equation is y= 1710. We can check our solution by substituting it into the original equation.
25=10y+8
Substitute

y= 17/10

25 ? = 10 ( 17/10 ) +8
â–Ľ
Simplify
DenomMultFracToNumber

10 * a/10= a

25 ? = 17+8
AddTerms

Add terms

25=25 âś“
Since the left-hand side is equal to the right-hand side, our solution is correct.
Subchapter links expand_more
arrow_right 1.2.1 Spatial Visualization and Reflections p.32-33
54
55
56
Spatial Visualization and Reflections 57 (Page 33)
58
arrow_right 1.2.2 Rigid Transformations: Rotations and Translations p.38
63
64
65
66
67
arrow_right 1.2.3 Slopes of Parallel and Perpendicular Lines p.42-43
Slopes of Parallel and Perpendicular Lines 76 (Page 42)
Slopes of Parallel and Perpendicular Lines 77 (Page 42)
78
Slopes of Parallel and Perpendicular Lines 79 (Page 43)
80
arrow_right 1.2.4 Defining Transformations p.48-49
Defining Transformations 85 (Page 48)
Defining Transformations 86 (Page 48)
Defining Transformations 87 (Page 48)
Defining Transformations 88 (Page 49)
89
arrow_right 1.2.5 Using Transformations to Create Shapes p.52-54
94
Using Transformations to Create Shapes 95 (Page 53)
96
Using Transformations to Create Shapes 97 (Page 53)
Using Transformations to Create Shapes 98 (Page 54)
arrow_right 1.2.6 Symmetry p.60-61
105
106
107
Symmetry 108 (Page 60)
Symmetry 109 (Page 61)