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 57: 92

Answer

$x=-4$ and $x=-\dfrac{7}{3}$

Work Step by Step

$\dfrac{x}{2x+7}-\dfrac{x+1}{x+3}=1$ Evaluate the substraction on the left: $\dfrac{x(x+3)-(x+1)(2x+7)}{(2x+7)(x+3)}=1$ $\dfrac{x^{2}+3x-[2x^{2}+9x+7]}{2x^{2}+13x+21}=1$ $\dfrac{x^{2}+3x-2x^{2}-9x-7}{2x^{2}+13x+21}=1$ $\dfrac{-x^{2}-6x-7}{2x^{2}+13x+21}=1$ Take the denominator to multiply the right side of the equation. We get: $-x^{2}-6x-7=2x^{2}+13x+21$ Take all terms to the right side of the equation: $2x^{2}+13x+21+x^{2}+6x+7=0$ Simplify: $3x^{2}+19x+28=0$ Solve by factoring: $(3x+7)(x+4)=0$ We get two solutions, which are: $x=-4$ and $x=-\dfrac{7}{3}$
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