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Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
Chapter Review
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Exercise 13 Page 153

Gather all of the variable terms on one side of the equation and all of the constant terms on the other.

-8

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other using the Properties of Equality. In this case, we start by subtracting 5 from both sides of the equation.

-3d/4+5=11
SubEqn

LHS-5=RHS-5

-3d/4+5-5=11-5
SubTerm

Subtract term

-3d/4=6
Now we can continue to solve using the Properties of Equality.

-3d/4=6
MoveNegFracToDenom

Put minus sign in denominator

3d/- 4=6
MultEqn

LHS * (-4)=RHS* (-4)

3d/- 4(-4)=6(-4)
MoveRightFacToNum

a/c* b = a* b/c

3d(-4)/- 4=6(-4)
CancelCommonFac

Cancel out common factors

3d=6(-4)
MultPosNeg

a(- b)=- a * b

3d=-24
DivEqn

.LHS /3.=.RHS /3.

3d/3=-24/3
CalcQuot

Calculate quotient

d=- 8

The solution to the equation is d=-8. We can check our solution by substituting it into the original equation.

-3d/4+5=11
Substitute

d= -8

-3( -8)/4+5 ? = 11
MultPosNeg

a(- b)=- a * b

-( -24)/4+5 ? = 11
MoveNegFracToNum

Put minus sign in numerator

-(-24)/4+5 ? = 11
NegNeg

- (- a)=a

24/4+5 ? = 11
CalcQuot

Calculate quotient

6+5 ? = 11
AddTerms

Add terms

11=11

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

Subchapter links expand_more
arrow_right Exercises p.152-156
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Exercises 7 (Page 153)
Exercises 8 (Page 153)
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