Question

How do you solve for x in `7/(x-2) + 3/(x+2) = -x/(x^2-4 )`?

Answer 1

Here is your answer `x= -8 / 11`

Explanation:

`(7(x+2) + 3 (x-2))/((x-2)(x+2))=-x/(x^2-4)`

`(7x+14+3x-6)/(x^2+2x-2x-4)=-x/(x^2-4)`

`(10x+8)/cancel(x^2-4)=-x/cancel(x^2-4)`

`10x + 8 = -x`

`11 x = -8`

`x= -8 / 11`

Answer 2

`x=-8/11`

Explanation:

Given equation:

`\frac{7}{x-2}+3/{x+2}=-x/{x^2-4}`

`\frac{7(x+2)+3(x-2)}{(x-2)(x+2)}=-x/{x^2-4}`

`\frac{10x+8}{x^2-4}=-x/{x^2-4}`

`\frac{10x+8}{x^2-4}+x/{x^2-4}=0`

`\frac{10x+8+x}{x^2-4}=0`

`\frac{11x+8}{x^2-4}=0`

`11x+8=0\ \quad(\forall \ \ x\ne\pm2 )`

`11x=-8`

`x=-8/11`

Answer 3

`x=-8/11`

Explanation:

It must be `x\ne 2,x\ne-2`
multimplying by `(x-2)(x+2)` we get

`7(x+2)+3(x-2)=-x` expanding we get

`7x+14+3x-6=-x`
combining like Terms

`10x+8=-x`
adding `-8` and `x`
We get

`11x=-8`

so `x=-8/11`

Answer 4

`x = -8/11`

Explanation:

`7/(x-2) + 3/(x+2) = -x/color(blue)((x^2-4 ))" "larr` factorise

`7/(x-2) + 3/(x+2) = -x/color(blue)((x+2)(x-2))`

Multiply each tern in the equation by the LCM of the denominators which is `color(magenta)((x+2)(x-2))` and cancel where possible:

`(color(magenta)((x+2)cancel((x-2)))xx7)/cancel((x-2)) + (3xxcolor(magenta)(cancel((x+2))(x-2)))/cancel((x+2)) = (-x xxcolor(magenta)cancel((x+2)(x-2)))/cancel((x+2)(x-2))`

This leaves:

`7(x+2)+3(x-2) =-x`

`7x+14 +3x-6 =-x`

`10x+x = -8`

`11x=-8`

`x =- 8/11`