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Pearson Algebra 1 Common Core, 2011

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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

LHS-5=RHS-5

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

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

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)

a(- b)=- a * b

3d=-24

.LHS /3.=.RHS /3.

3d/3=-24/3

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

d= -8

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

a(- b)=- a * b

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

MoveNegFracToNum

Put minus sign in numerator

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

- (- a)=a

24/4+5 ? = 11

Calculate quotient

6+5 ? = 11

Add terms

11=11

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

Subchapter links

Exercises p.152-156