Question

Solve
\[(x/x-2)+(3/x+3)=(-14/x^2+x-6)\]

Answer 1

if i tell you that x^2 + x - 6 = (x-2)(x+3)

will that help?

Answer 2

Parentheses in the right places? Or does it look like this:
\[\frac{x}{x-2}+\frac{3}{x+3}=\frac{-14}{x^2+x-6}\]

Answer 3

it looks that that

Answer 4

Multiply each term by the common denominator which is (x-2)(x+3)

Answer 5

x(x+3)+3(x-2)=-14

Answer 6

Take it from there.

Answer 7

im confused

Answer 8

so x-2 i only multiply the x+3

Answer 9

i only multiply the missing ones?

Answer 10

When you multiply every term by (x-2)(x+3), what do you get?

Answer 11

But aren't you supposed to cancel out terms that are already there?

Answer 12

Right :) . What do you get after you do that?

Answer 13

\[\frac{x}{x-2}\times(x-2)(x+3)+\frac{3}{x+3}\times(x-2)(x+3)=\frac{-14}{(x-2)(x+3})\times(x-2)(x+3)\]

Answer 14

\[x(x+3)+3(x-2)=-14\]