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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 25 Page 34

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

No solution.

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-hand side of the equation.

1/5(15b-7) = 3b - 9

Distribute 1/5

1/5(15b) - 1/5(7)= 3b - 9

MoveRightFacToNumOne

1/b* a = a/b

15b/5 - 7/5 = 3b - 9

Calculate quotient

3b - 7/5 = 3b - 9

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

3b - 7/5 = 3b - 9

LHS-3b=RHS-3b

3b - 7/5 - 3b = 3b - 9 - 3b

CommutativePropAdd

Commutative Property of Addition

3b - 3b - 7/5 = 3b - 3b - 9

Subtract terms

- 7/5 ≠ - 9 *

Simplifying the equation resulted in a contradiction. Therefore, the equation has no solution.

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