Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 38

Answer

$a = \frac{b^2+b}{2}$

Work Step by Step

$Solve$ $the$ $equation$ $for$ $the$ $indicated$ $variable:$ $\frac{a+1}{b} = \frac{a-1}{b}+ \frac{b+1}{a};$ $for$ $a$ Solve for a Multiply both sides by $ab$ [Note: We combine the steps for multiply a and b for both sides to speed up the process] $ ab(\frac{a+1}{b}) = ab(\frac{a-1}{b}+ \frac{b+1}{a})$ $ab(\frac{a+1}{b}) = ab(\frac{a-1}{b})+ ab(\frac{b+1}{a})$ Simplify $a(a+1) = a(a-1)+ b(b+1)$ $a^2 + a = a^2-a + b^2 + b$ Subtract $a^2$ from both sides $a^2 + a - a^2 = a^2 -a + b^2 + b - a^2$ Simplify $a = -a + b^2 + b$ Add $a$ to both sides $a+a = -a + b^2 + b + a$ Simplify $2a = b^2 + b$ Divide both sides by 2 $\frac{2a}{2} = \frac{b^2+b}{2}$ $a = \frac{b^2+b}{2}$
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