How do you solve $3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4)$ and check for extraneous solutions?

Question

2129 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T23:46:11

Solution is $x=1$

$3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4)$

$hArr(3(x+4)-1(x+1))/(x^2+5x+4)=(x+12)/(x^2+5x+4)$ or

$(3x+12-x-1)/(x^2+5x+4)=(x+12)/(x^2+5x+4)$ or

$(2x+11)/(x^2+5x+4)=(x+12)/(x^2+5x+4)$

Now assuming $x^2+5x+4!=0$ (i.e. $x$ is neither equal to $-1$ nor equal to $-4$ ), and multiplying each side by $x^2+5x+4$ , we get

$2x+11=x+12$ or

$x=1$

As $x$ is neither equal to $-1$ nor equal to $-4$ , there is no extraneous solution.

How do you solve 3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4)3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4) and check for extraneous solutions?