Question

Subtract$\frac{6c}{c^2-25}-\frac{3}{c+5}$.

Short answer

The result is$\frac{3}{c-5}$.

Step-by-step solution

Given Information   

The Given expression is$\frac{6c}{c^2-25}-\frac{3}{c+5}$.

Find the LCD of denominators.   

Factor the denominators.

$c^2-25=(c+5)(c-5)$

$c+5=(c+5)$

LCD$=(c+5)(c-5)$

Rewrite the expression in factored form.

$=\frac{6c}{(c+5)(c-5)}-\frac{3}{(c+5)}$

Rewrite each fraction as an equivalent fraction with the LCD.   

$=\frac{6c}{(c+5)(c-5)}-\frac{3(c-5)}{(c+5)(c-5)}$

Simplify numerators.

$=\frac{6c}{(c+5)(c-5)}-\frac{3c-15}{(c+5)(c-5)}$

Subtract and simplify the numerator.  

Take common denominator.

$=\frac{6c-(3c-15)}{(c+5)(c-5)}$

$=\frac{6c-3c+15}{(c+5)(c-5)}$

Simplify numerator.

$=\frac{3c+15}{(c+5)(c-5)}$

$=\frac{3(c+5)}{(c+5)(c-5)}$

$=\frac{3}{c-5}$