# Can someone please explain how to solve for x? Thank you!

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To solve: `3/(1+2x)-2/(1-2x)=21/(1-4x^2)`

`rArr (3(1-2x)-2(1+2x))/((1+2x)(1-2x))=21/(1-4x^2)`

Applying the formula `(a+b)(a-b)=a^2-b^2 ` we get:

`(3(1-2x)-2(1+2x))/(1-4x^2)=21/(1-4x^2)`

Multiplying both sides by `(1-4x^2)` :

`3(1-2x)-2(1+2x)=21`

`rArr 3-6x-2-4x=21`

`rArr 1-10x=21`

`rArr x=-20/10=-2`

**Therefore, the value of x is -2.**

You need to notice that converting the right side denominator into a product, yields:

`1 - 4x^2 = ((1 - 2x)(1 + 2x))`

You need to bring the fractions to a common denominator that is `1 - 4x^2,` hence, you need to multiply the first fraction to the left by `1 - 2x` and the second fraction to the left by `1 + 2x` , such that:

`(3(1 - 2x) -2(1 + 2x))/((1 - 2x)(1 + 2x)) = 21/((1 - 2x)(1 + 2x))`

`3 - 6x - 2 - 4x = 21`

You need to isolate the terms that contain x to the left side, such that:

`-10x = 21 - 1 => -10x = 20 => x = 20/(-10) => x = -2`

**Hence, evaluating the solution to the given equation, yields x = -2**.