Question

8/x-5 - 9/x-4 = 5/x^2-9x+20

Answer 1

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Solve the equation:

$\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}$


Reduce the fractions at the left side so that they have the same denominator:

$\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\ \mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\ \mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\ \mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}$


Numerators must be equal:

$\mathsf{8x-32-(9x-45)=5}\\\\ \mathsf{8x-32-9x+45=5}\\\\ \mathsf{8x-9x=5+32-45}\\\\ \mathsf{-x=-8}\\\\ \mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}$


I hope this helps. =)


Tags:  rational equation fraction solution algebra