How do you solve $3/(x+4)-1/(x+3)=(x+9)/(x^2+7x+12)$ and check for extraneous solutions?

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Answer 1 · 2016-12-30T09:55:08

$x=4$

Given:

$3/(x+4)-1/(x+3) = (x+9)/(x^2+7x+12)$

Multiply both sides by:

$(x+4)(x+3) = x^2+7x+12$

to get:

$3(x+3)-1(x+4) = x+9$

Expanding, this becomes:

$3x+9-x-4 = x+9$

Simplifying:

$2x+5 = x+9$

Subtract $x+5$ from both sides to get:

$x = 4$

This solution is valid since $(color(blue)(4)+4) = 8 != 0$ and $(color(blue)(4)+3) = 7 != 0$ .

How do you solve 3/(x+4)-1/(x+3)=(x+9)/(x^2+7x+12)3/(x+4)-1/(x+3)=(x+9)/(x^2+7x+12) and check for extraneous solutions?