## Precalculus (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises: 32

#### Answer

The solutions are $x=\dfrac{4}{3}$ and $x=\dfrac{9}{4}$

#### Work Step by Step

$6=\dfrac{7}{2x-3}+\dfrac{3}{(2x-3)^{2}}$ Multiply the whole equation by $(2x-3)^{2}$: $(2x-3)^{2}\Big[6=\dfrac{7}{2x-3}+\dfrac{3}{(2x-3)^{2}}\Big]$ $6(2x-3)^{2}=7(2x-3)+3$ Evaluate the indicated operations: $6(4x^{2}-12x+9)=14x-21+3$ $24x^{2}-72x+54=14x-21+3$ Take all terms to the left side and simplify: $24x^{2}-72x+54-14x+21-3=0$ $24x^{2}-86x+72=0$ Divide the whole equation by $2$: $\dfrac{1}{2}(24x^{2}-86x+72=0)$ $12x^{2}-43x+36=0$ Solve by factoring: $(3x-4)(4x-9)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $3x-4=0$ $3x=4$ $x=\dfrac{4}{3}$ $4x-9=0$ $4x=9$ $x=\dfrac{9}{4}$ The solutions are $x=\dfrac{4}{3}$ and $x=\dfrac{9}{4}$

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