Question

(11/x^2-25)+(4/x+5)= 3/x-5

Answer 1

$\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -------------------------------\\\\ \cfrac{11}{x^2-25}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\implies \cfrac{11}{x^2-5^2}+\cfrac{4}{x+5}=\cfrac{3}{x-5}$

$\bf \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\impliedby \begin{array}{llll} \textit{notice, LCD is }(x-5)(x+5)\\ \textit{so let's multiply all by it}\\ \textit{to toss the denominators} \end{array}$

$\bf (x-5)(x+5)\left( \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5} \right)=(x-5)(x+5)\left( \cfrac{3}{x-5} \right) \\\\\\ 11+4(x-5)=3(x+5)\implies 11+4x-20=3x+15 \\\\\\ 4x-9=3x+15\implies x=24$