Question

How do you solve `\frac { 4} { x - 3} + \frac { 3} { x + 9} = \frac { 6} { ( x + 9) ( x - 3) }`?

Answer 1

`x=-3`

But, `x` cannot equal to `3` nor `-9`.

Explanation:

Let's start with the original problem:

`4/(x-3)+3/(x+9)=6/((x+9)(x-3))`

Seeing as how there are fractions, we can multiply by both sides `(x-3)(x+9)`. This has the purpose of removing the `(x-3)` and the `(x+9)` on the left side and the `(x-3)(x+9)` on the right side:

`(x-3)(x+9)(4/(x-3)+3/(x+9))=(x-3)(x+9)(6/((x+9)(x-3)))`

Then we can use the Distributive Property to simplify out the left side:

`(x-3)(x+9)(4/(x-3))+(x-3)(x+9)(3/(x+9))=(x-3)(x+9)(6/((x+9)(x-3)))`

Now we can cancel out some terms and then simplify the result:

`cancel(x-3)(x+9)(4/cancel(x-3))+(x-3)cancel(x+9)(3/cancel(x+9))=cancel((x-3)(x+9))(6/cancel((x+9)(x-3)))`

`4(x+9)+3(x-3)=6`

We can use the Distributive Property again and simplify the equation:

`4x+36+3x-9=6`

We then combine like terms and move all the `x` terms to the left and the numerical terms to the right:

`7x+27=6`

`7x=-21`

We then divide both sides by `7` to isolate `x`:

`x=-3`

However, we're not done. We have to make sure that the value of our variable won't make any of the denominators in the original equation turn to `0` because otherwise, the solution will be an extraneous solution.

So, when `x=-3`,

• `x-3=-3-3=color(red)(-6)`
• `x+9=-3+9=color(red)6`
• `(x+9)(x-3)=(-3+9)(-3-3)=(6)(-6)=color(red)(-36)`

None of these denominators are equal to `0` so therefore, our solution is `x=-3`.

However, there are some solutions that do make the denominators `0`. For example:

• For the denominator `x-3`, if `color(blue)(x=3)`, then the denominator will turn to `0`, making `3` a value that `x` cannot be.

• For the denominator `x+9`, if `color(blue)(x=-9)`, then the denominator will turn to `0`, making `-9` a value that `x` cannot be.

• For the denominator `(x-3)(x+9)`, if `color(blue)(x=3)` or `color(blue)(x=-9)`, then the denominator will turn to `0`, making `3` and `-9` values that `x` cannot be.

In conclusion, `x` cannot be `3` nor `-9`, making these two values values `x` cannot be.

Hope this helped!