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

May 28, 2018

$x = - 1 \frac{1}{2} \vee x = 2$

## 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 ${x}^{2} + b x + c = 0$.
If you find two values whose sum is $b$ and whose product is $c$, you can find $x$ because
${x}^{2} + \left(\text{value1")x + ("value2")x + ("value1")("value2}\right) = 0$

$\left(x + \text{value1")(x+"value2}\right) = 0$

$x + \text{value1"=0 vv x+"value2} = 0$.

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

$\frac{6 x}{x + 4} + 4 = \frac{2 x + 2}{x - 1}$

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

$6 x + 4 \left(x + 4\right) = \frac{2 x + 2}{x - 1} \left(x + 4\right)$

$10 x + 16 = \frac{2 x + 2}{x - 1} \left(x + 4\right)$

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

$\left(10 x + 16\right) \left(x - 1\right) = \left(2 x + 2\right) \left(x + 4\right)$

$10 {x}^{2} - 10 x + 16 x - 16 = 2 {x}^{2} + 8 x + 2 x + 8$

$10 {x}^{2} + 6 x - 16 = 2 {x}^{2} + 10 x + 8$

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

$8 {x}^{2} - 4 x - 24 = 0$

Divide both sides by 8.

${x}^{2} - \frac{1}{2} x - 3 = 0$

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

The numbers $1 \frac{1}{2}$ and $- 2$ work, because

$1 \frac{1}{2} + \left(- 2\right) = - \frac{1}{2} = b$ and

$1 \frac{1}{2} \cdot - 2 = - 3 = c$

So,

$\left(x + 1 \frac{1}{2}\right) \left(x - 2\right) = 0$

$x + 1 \frac{1}{2} = 0 \vee x - 2 = 0$

$x = - 1 \frac{1}{2} \vee x = 2$

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