How do you solve $1/(x-4) + (x-4)/(x-4) = 7 / ( x^2+x-20)$ ?

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Answer 1 · 2017-06-19T17:26:16

Eliminate the denominators by multiplying both sides by $(x-4)(x+5)$ .
Solve the resulting quadratic.

Given: $1/(x-4) + (x-4)/(x-4) = 7 / ( x^2+x-20)$

Eliminate the denominators by multiplying both sides by $(x-4)(x+5)$ .

$x+5+(x-4)(x+5) = 7$

$x+5 + x^2+x-20 = 7$

$x^2+2x-22 = 0$

$x = (-2+-sqrt(2^2-4(1)(-22)))/(2(1))$

$x = (-2+-2sqrt(23))/2$

$x = -1+sqrt(23) and x = -1-sqrt(23)$

This agrees with WolframAlpha

How do you solve 1/(x-4) + (x-4)/(x-4) = 7 / ( x^2+x-20)1/(x-4) + (x-4)/(x-4) = 7 / ( x^2+x-20)?