Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2: 52

Answer

y = 1

Work Step by Step

$\frac{y}{12}$ + $\frac{1}{6}$ = $\frac{y}{2}$ - $\frac{1}{4}$ First, we can do away with the fractions if we find the least common denominator and multiply each term of the equation by that number. We will write the number over one so that it is in fraction form. The least common denominator for the problem is 12, so we will multiply every term by $\frac{12}{1}$. Multiply every term by $\frac{12}{1}$ ($\frac{12}{1}$)($\frac{y}{12}$) + ($\frac{12}{1}$)($\frac{1}{6}$) = ($\frac{12}{1}$)($\frac{y}{2}$) - ($\frac{12}{1}$)($\frac{1}{4}$) Completing the multiplication, we getL $\frac{12y}{12}$ + $\frac{12}{6}$ = $\frac{12y}{2}$ - $\frac{12}{4}$ We can now simplify each fraction to get: y + 2 = 6y - 3 Subtract 6y from both sides of the equation. y - 6y + 2 = 6y - 6y - 3 Complete the arithmetic to get: -5y + 2 = -3 Now, subtract 2 from each side. -5y + 2 - 2 = -3 - 2 Complete the arithmetic. -5y = -5 Divide both sides by -5. $\frac{-5y}{-5}$ = $\frac{-5}{-5}$ Complete the arithmetic and get: y = 1 Checking our answer by substituting 1 back in to the original equation for y. $\frac{1}{12}$ + $\frac{1}{6}$ = $\frac{1}{2}$ - $\frac{1}{4}$ Rewrite each fraction as an equivalent fraction using the least common denominator of 12. $\frac{1}{12}$ + $\frac{2}{12}$ = $\frac{6}{12}$ - $\frac{3}{12}$ Perform the arithmetic and we get $\frac{3}{12}$ = $\frac{3}{12}$ Since out last line matches, we know that our solution of y = 1 is correct.
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