## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 2 - Section 2.4 - Formulas and Percents - Exercise Set - Page 155: 85

#### Answer

The solution is: $$y = 20$$ To determine if the solution is correct, we plug $20$ in for $y$ into the equation to see if both sides of the equation are equal: $$5(2(20) - 3) - 1 = 4(6 + 2(20))$$ We can simplify what we have in parentheses: $$5(40 - 3) - 1 = 4(6 + 40)$$ $$5(37) - 1 = 4(46)$$ We simplify using order of operations to multiply first: $$185 - 1 = 184$$ We then do the subtraction: $$184 = 184$$ We see that the solution is correct because both sides of the equation are equal.

#### Work Step by Step

To solve, we have to simplify the equation first. We distribute the terms in parentheses: $$10y - 15 - 1 = 24 + 8y$$ We now move the $y$ terms to one side of the equation by subtracting $8y$ from both sides of the equation: $$2y - 15 - 1 = 24$$ We want constants on the other side of the equation, so we add both $15$ and $1$ to both sides of the equation: $$2y = 40$$ We now isolate $y$ by dividing both sides by $2$: $$y = 20$$ To determine if the solution is correct, we plug $20$ in for $y$ into the equation to see if both sides of the equation are equal: $$5(2(20) - 3) - 1 = 4(6 + 2(20))$$ We can simplify what we have in parentheses: $$5(40 - 3) - 1 = 4(6 + 40)$$ $$5(37) - 1 = 4(46)$$ We simplify using order of operations to multiply first: $$185 - 1 = 184$$ We then do the subtraction: $$184 = 184$$ We see that the solution is correct because both sides of the equation are equal.

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