How do you solve `(6x) /(x+4) + 4 = (2x + 2) /(x - 1)`?

Information

In equations, you are allowed to perform operations on both sides. In order to get rid of denominators, you can multiply by the denominator on both sides. Since a number divided by itself is 1, except for 0, this removes the fraction and leaves the numerator.

The product-sum method can be used to find values of `x` when the equation is shaped like `x^2+bx+c=0`.
If you find two values whose sum is `b` and whose product is `c`, you can find `x` because
`x^2 + ("value1")x + ("value2")x + ("value1")("value2")=0`

`(x+"value1")(x+"value2")=0`

`x+"value1"=0 vv x+"value2"=0`.

Note that you can use the quadratic formula in all cases where the product-sum method is applicable, but it is much faster in a lot of cases to use this.

Calculation

`(6x)/(x+4)+4=(2x+2)/(x-1)`

Multiply both sides by `x+4` to get rid of the denominator on the left side.

`6x+4(x+4)=(2x+2)/(x-1)(x+4)`

`10x+16=(2x+2)/(x-1)(x+4)`

Multiply both sides by `x-1` to do the same for the right side.

`(10x+16)(x-1)=(2x+2)(x+4)`

`10x^2-10x+16x-16=2x^2+8x+2x+8`

`10x^2+6x-16=2x^2+10x+8`

Make the right side equal to 0 by subtracting everything on the right side from both sides.

`8x^2-4x-24=0`

Divide both sides by 8.

`x^2-1/2x-3=0`

Use the product-sum method to find `x`.

The numbers `1 1/2` and `-2` work, because

`1 1/2+(-2)=-1/2=b` and

`1 1/2*-2=-3=c`

So,

`(x+1 1/2)(x-2)=0`

`x+1 1/2=0 vv x-2=0`

`x=-1 1/2 vv x=2`

Neither of these values makes a division by zero when filled back into the equation, so they're both valid.