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

Feb 13, 2016

Put it on a common denominator.

#### Explanation:

The LCD (Least Common Denominator) would be (x + 4)(x - 1)

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

We can now eliminate the denominators, because both fractions are now equivalent.

$6 {x}^{2} + 4 x - 6 x - 4 = 2 {x}^{2} + 2 x + 8 x + 8$

Since this is a quadratic equation, we must put everything to one side of the equation, so that the other side is 0.

$4 {x}^{2} - 12 x - 12 = 0$

$4 \left({x}^{2} - 3 x - 3 x\right) = 0$

As you can see, this cannot be factored. I will solve by completing the square, but you could also use the quadratic formula.

$4 \left({x}^{2} - 3 x + n\right) = 12$

$n = {\left(\frac{b}{2}\right)}^{2}$

$n = {\left(- \frac{3}{2}\right)}^{2}$

$n = \frac{9}{4}$

$4 \left({x}^{2} - 3 x + \frac{9}{4} - \frac{9}{4}\right) = 12$

$4 \left({x}^{2} - 3 x + \frac{9}{4}\right) - 4 \left(\frac{9}{4}\right) = 12$

$4 {\left(x - \frac{3}{2}\right)}^{2} - 9 = 12$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{21}{4}$

$\left(x - \frac{3}{2}\right) = \pm \sqrt{\frac{21}{4}}$

$x = \frac{\pm \sqrt{21}}{2} + \frac{3}{2}$

$x = \frac{\pm \sqrt{21} + 3}{2}$