Question

How do you solve `7/(x-4) = 1 + 9/(x+4)`?

Answer 1

`x = -8` or `x = 10`

Explanation:

first, a common denominator for all the fractions needs to be found.

this is the lowest common multiple of `1`, `x - 4`, and `x + 4`.

since `x + 4` and `x - 4` have no common factors, their lowest common multiple is the two expressions multiplied together.

this gives `(x+4)(x-4)`.

`(x+4)(x-4)` can be used as the denominator for both fractions:

`(7)/(x-4) = (7(x+4))/((x-4)(x+4))`

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

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

`(7)/(x-4) = 1 + (9)/(x+4)`

`(7(x+4))/((x-4)(x+4)) = ((x-4)(x+4))/((x-4)(x+4)) + (9(x-4))/((x-4)(x+4))`

the denominator can then be taken away by multiplying everything by `(x-4)(x+4):`

`7(x+4)= (x-4)(x+4) + 9(x-4)`

then the brackets can be expanded:

`7(x+4) = 7x + 28`

`(x-4)(x+4) = x^2 - 16`

(difference of two squares identity: `(a+b)(a-b) = a^2-b^2`)

`9(x-4) = 9x - 36`

`(7x + 28) = (x^2-16) + (9x-36)`

`7x+28 = x^2-16 + 9x - 36`

collect like terms:

`7x + 28 = x^2 + 9x - 16 - 36`

`7x + 28 = x^2 + 9x - 52`

`x^2 + 9x - 52 = 7x + 28`

then subtract `7x:`

`x^2 + 2x - 52 = 28`

subtract `28` so that `0` is on the right-hand side:

`x^2 - 2x - 80 = 0`

this forms a quadratic eqaution that can be factorised.

to do this, find two numbers that add to make `-2`

and multiply to make `-80`.

`8 + (- 10) = 8 - 10 = -2`
`8 * -10 = -80`

`x^2 - 2x - 80 = (x+8)(x-10)`

`(x+8)(x-10) = 0`

for the number on the right-hand side to be `0`, at least one of the numbers in brackets has to be `0`.

either `x+8 = 0` or `x - 10 = 0`

if `x + 8 = 0`, then `x = -8`.

if `x - 10 = 0`, then `x = 10`.

this means that the two possible values for `x` are `-8` and `10`.

Answer 2

`x` equals `8 and -10`

Explanation:

`7/(x−4)=1/(1)+9/(x+4)`     Solve for `x`

1) Give the addends on the right side a common denominator

`7/(x−4)=(x+4)/(x+4)+9/(x+4)`

2) Add the like fractions on the right by adding the numerators and keeping the common denominator

`7/(x−4)=(x+4+9)/(x+4)`

3) Combine like terms in the numerator on the right

`7/(x−4)=(x+13)/(x+4)`

4) Clear the first fraction by multiplying both sides by `x - 4` and letting the denominator on the left cancel

`7=((x-4)(x+13))/(x+4)`

5) Clear the second fraction by multiplying both sides by `x+4` and letting the right side denominator cancel

`7(x + 4)=(x−4)(x+13)`

6) Clear the parentheses by distributing the `7` and by multiplying the binomials

`7x + 28 = x^2 +9x - 52`

7) Subtract `7x` from both sides to collect all the `x` terms together

`28 = x^2 + 2x - 52`

8) Set the equation to `0` by subtracting `28` from both sides

`x^2 + 2x - 80 = 0`

9) Factor

`(x + 10)(x - 8) = 0`

10) Set the factors equal to `0` and solve for `x`

`x + 10 = 0`
`x = -10`

`x - 8 = 0`
`x = 8`

Answer:
`(8,-10)`

`color(white)(mmmmmmmm)`―――――――――

Check

Sub in `8` in the original equation to be sure it still is an equality
`7/(x−4)=1/(1)+9/(x+4)`

`7/(8−4)` should equal `1/(1)+9/(8+4)`

`7/(4)` should still equal `12/(12)+9/(12)`

`7/(4)` should still equal `21/(12)`

Reduce the fraction to lowest terms

`7/4` does equal `7/4`

`Check`