Question

Write as equivalent rational expression with LCD $(a+9)(a+5)(a-9)$of $\frac{6}{a^2+14a+45},\frac{5a}{a^2-81}$.

Short answer

The result is$\frac{6a-54}{(a+9)(a+5)(a-9)},\frac{5a^2+25a}{(a+9)(a-9)(a+5)}$.

Step-by-step solution

Given Information

The given expression is$\frac{6}{a^2+14a+45},\frac{5a}{a^2-81}$

The LCD is$(a+9)(a+5)(a-9)$.

Factor the denominator of both expressions.

The factored form of first denominator.

$a^2+14a+45=(a+9)(a+5)$

The factored form of second denominator.

$a^2-81=(a+9)(a-9)$

The fractions in factored form is$\frac{6}{(a+9)(a+5)},\frac{5a}{(a+9)(a-9)}$.

Multiply each denominator by the 'missing' factor and multiplyeach numerator by the same factor.

$\frac{6(a-9)}{(a+9)(a+5)(a-9)},\frac{5a(a+5)}{(a+9)(a-9)(a+5)}$

Simplify the numerators.

Multiply the numerator using distributive property.

$\frac{6a-54}{(a+9)(a+5)(a-9)},\frac{5a^2+25a}{(a+9)(a-9)(a+5)}$