# Math please help! Solve algebraically for x: `1/(x+3) - 2/(3-x) = 4/(x^2-9 )`

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`1/(x+3) - 2/(3-x) = 4/(x^2-9)`

Factor out the negative in 3-x.

`1/(x+3) - 2/(-(x-3)) = 4/(x^2-9)`

`1/(x+3) + 2/(x-3) = 4/(x^2-9)`

Next, factor the quadratic expression `x^2-9` .

`1/(x+3) + 2/(x-3) = 4/((x+3)(x-3))`

To eliminate the denominators, multiply both sides by the LCD which is (x+3)(x-3).

`(x+3)(x-3)*(1/(x+3) + 2/(x-3)) = (4/((x+3)(x-3)))*(x+3)(x-3)`

`x-3 + 2(x+3)=4`

`x-3+2x+6=4`

At the left side of the equation, combine like terms.

`3x+3=4`

Then, isolate x. So subtract both sides by 3.

`3x+3-3=4-3`

`3x=1`

And divide both sides by 3.

`(3x)/3 = 1/3`

`x=1/3` **Hence, the solution to the given rational equation is `x=1/3` .**

we know that;

`(a+b)(a-b) = a^2-b^2`

`1/(x+3)-2/(3-x) = 4/(x^2-9)`

`(3-x-2(x+3))/((x-3)(3-x)) = 4/(x^2-9)`

`(-3x-3)/((x+3)(-1)(x-3)) = 4/(x^2-9)`

`((-)(-3x-3))/((x-3)(x+3)) = 4/(x^2-9)`

` (3(x+1))/(x^2-9) = 4/(x^2-9)`

`3(x+1) = 4`

` x = 4/3-1`

` x = 1/3`

*So the answer is x = 1/3*

We kno that

`(a+b)(a-b) = a^2-b^2`

` 1/(x+3)-2/(3-x) = 4/(x^2-9)`

`(3-x-2(x+3))/((x-3)(3-x)) = 4/(x^2-9)`

` (-3x-3)/((x+3)(-1)(x-3)) = 4/(x^2-9)`

` ((-)(-3x-3))/((x-3)(x+3)) = 4/(x^2-9)`

`(3(x+1))/(x^2-9) = 4/(x^2-9)`

` 3(x+1) = 4`

`x = 4/3-1`

`x = 1/3`

*So the answer is x = 1/3*