Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 9 - Sequences and Series - 9-1 Mathematical Patterns - Practice and Problem-Solving Exercises - Page 571: 76

Answer

$y = 2$

Work Step by Step

Before we can solve the equation, find the least common denominator for the two fractions. The least common denominator, or LCD, is $3(y + 1)$, in this case. Convert each fraction to an equivalent one by multiplying its numerator with whatever factor is missing between its denominator and the LCD: $\dfrac{y(3)}{3(y + 1)} = \dfrac{2(y + 1)}{3(y + 1)}$ Distribute and multiply to simplify: $\dfrac{3y}{3(y + 1)} = \dfrac{2y + 2}{3(y + 1)}$ Multiply both sides of the equation by $3(y + 1)$ to eliminate the fractions: $3y = 2y + 2$ Subtract $2y$ from each side of the equation: $y = 2$ To check the solution, plug in $2$ for $y$ into the original equation: $\dfrac{2}{2 + 1} = \dfrac{2}{3}$ Simplify: $\frac{2}{3} = \frac{2}{3}$ Both sides are equal to one another; therefore, this solution is correct.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.