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Big Ideas Math: Modeling Real Life, Grade 8

BI

Big Ideas Math: Modeling Real Life, Grade 8 View details

Chapter Review

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Exercise 22 Page 34

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

p = 0.4

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 side using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the left- and right-hand sides of the equation.

3(5p - 3)= 5(p - 1)

Distribute 3

15p - 9 = 5(p - 1)

Distribute 5

15p - 9 = 5p - 5

Now we can continue to solve using the Properties of Equality.

15p - 9 = 5p - 5

LHS-5p=RHS-5p

15p - 9 - 5p = 5p - 5 - 5p

CommutativePropAdd

Commutative Property of Addition

15p - 5p - 9 = 5p - 5p - 5

Subtract term

10p - 9 = - 5

LHS+9=RHS+9

10p - 9 + 9 = - 5 + 9

Add terms

10p = 4

.LHS /10.=.RHS /10.

p = 0.4

The solution to the equation is p=0.4. We can check our solution by substituting it into the original equation.

3(5p - 3) = 5(p - 1)

p= 0.4

3(5( 0.4)-3) ? = 5( 0.4 - 1)

Multiply

3(2 - 3) ? = 5(0.4 - 1)

Subtract terms

3(- 1)? =5(- 0.6)

Multiply

- 3 = - 3 ✓

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

Subchapter links

1 Solving Simple Equations p.33

2 Solving Multi-Step Equations p.34

3 Solving Equations with Variables on Both Sides p.34-35

4 Rewriting Equations and Formulas p.35