Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 7 - Section 7.5 - Solving Equations Containing Rational Expressions - Exercise Set: 42


$t=-\dfrac{1}{2}$ and $t=-6$

Work Step by Step

$\dfrac{2t+3}{t-1}-\dfrac{2}{t+3}=\dfrac{5-6t}{t^{2}+2t-3}$ Factor the denominator of the third fraction: $\dfrac{2t+3}{t-1}-\dfrac{2}{t+3}=\dfrac{5-6t}{(t+3)(t-1)}$ Multiply the whole equation by $(t+3)(t-1)$: $(t+3)(t-1)\Big[\dfrac{2t+3}{t-1}-\dfrac{2}{t+3}=\dfrac{5-6t}{(t+3)(t-1)}\Big]$ $(2t+3)(t+3)-2(t-1)=5-6t$ $2t^{2}+9t+9-2t+2=5-6t$ Take all terms to the left side of the equation and simplify it by combining like terms: $2t^{2}+9t+9-2t+2-5+6t=0$ $2t^{2}+13t+6=0$ Solve this equation by factoring: $(2t+1)(t+6)=0$ The two solutions are: $t=-\dfrac{1}{2}$ and $t=-6$ The original equation is not undefined for neither of these values found. The answer is: $t=-\dfrac{1}{2}$ and $t=-6$
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