What is x/(x^2+9x+20)-4/(x^2+7x+12) ?

The solution to x/(x^2+9x+20)-4/(x^2+7x+12) is (x-5)/((x+5)(x+3))

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simplify x/(x^2+9x+20)-4/(x^2+7x+12)

simplify

\frac{x}{x ^{2} + 9 x + 20} - \frac{4}{x ^{2} + 7 x + 12}

\frac{x - 5}{( x + 5 ) ( x + 3 )}

Solution steps

\frac{x}{x ^{2} + 9 x + 20} - \frac{4}{x ^{2} + 7 x + 12}

Least Common Multiplier of

x^{2} + 9 x + 20, x^{2} + 7 x + 12 : (x + 4) (x + 5) (x + 3)

Adjust Fractions based on the LCM

= \frac{x ( x + 3 )}{( x + 4 ) ( x + 5 ) ( x + 3 )} - \frac{4 ( x + 5 )}{( x + 4 ) ( x + 5 ) ( x + 3 )}

Apply the fraction rule:

\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}

= \frac{x ( x + 3 ) - 4 ( x + 5 )}{( x + 4 ) ( x + 5 ) ( x + 3 )}

Simplify

x (x + 3) - 4 (x + 5) : x^{2} - x - 20

= \frac{x ^{2} - x - 20}{( x + 4 ) ( x + 5 ) ( x + 3 )}

Cancel

\frac{x ^{2} - x - 20}{( x + 4 ) ( x + 5 ) ( x + 3 )} : \frac{x - 5}{( x + 5 ) ( x + 3 )}

= \frac{x - 5}{( x + 5 ) ( x + 3 )}

(x - 1) (x^{2} - 1)

\frac{16}{60}

\frac{1}{1 + tan ^{2} ( x )} + \frac{1}{1 + cot ^{2} ( x )}

\frac{5}{x + 9} + \frac{5}{x}

4 y

What is x/(x^2+9x+20)-4/(x^2+7x+12) ?
The solution to x/(x^2+9x+20)-4/(x^2+7x+12) is (x-5)/((x+5)(x+3))