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

Dec 23, 2016

The solution for this equation is $x = 2 , - \frac{3}{2}$

Explanation:

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

or, $\frac{6 x + 4 \cdot \left(x + 4\right)}{x + 4} = \frac{2 x + 2}{x - 1} \rightarrow$ Take LCM on LHS.

or, $\frac{6 x + 4 x + 16}{x + 4} = \frac{2 x + 2}{x - 1}$

or, $\frac{10 x + 16}{x + 4} = \frac{2 x + 2}{x - 1}$

or, $\left(x - 1\right) \left(10 x + 16\right) = \left(x + 4\right) \left(2 x + 2\right) \rightarrow$ Cross Multiply

or, $x \left(10 x + 16\right) - 1 \left(10 x + 16\right) = x \left(2 x + 2\right) + 4 \left(2 x + 2\right) \rightarrow$Simplify

or, $10 {x}^{2} + 16 x - 10 x - 16 = 2 {x}^{2} + 2 x + 8 x + 8 \rightarrow$Simplify

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

or, $2 \left(5 {x}^{2} + 3 x - 8\right) = 2 \left({x}^{2} + 5 x + 4\right) \rightarrow$Take 2 Common on both LHS Side and RHS Side

or, $\frac{2 \left(5 {x}^{2} + 3 x - 8\right)}{2} = \frac{2 \left({x}^{2} + 5 x + 4\right)}{2} \rightarrow$Dividing both sides by 2

or, $5 {x}^{2} + 3 x - 8 = {x}^{2} + 5 x + 4$

or, $5 {x}^{2} - {x}^{2} + 3 x - 3 x - 8 = {x}^{2} - {x}^{2} + 5 x - 3 x + 4 \rightarrow$Subracting ${x}^{2}$ and 3x from boths sides

or, $4 {x}^{2} - 8 = 2 x + 4$

or, $4 {x}^{2} - 8 - 4 = 2 x + 4 - 4 \rightarrow$Subtracting 4 from both sides

or, $4 {x}^{2} - 12 = 2 x$

or, $4 {x}^{2} - 2 x - 12 = 2 x - 2 x \rightarrow$ Subtracting 2x from both sides

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

or, $4 {x}^{2} - 8 x + 6 x - 12 = 0 \rightarrow$ Factoring LHS

or, $4 x \left(x - 2\right) + 6 \left(x - 2\right) = 0$

or, $\left(4 x + 6\right) \left(x - 2\right) = 0$

Either 4x+6=0
or, $4 x = - 6$

or, $x = - \left(\frac{6}{4}\right)$

or, $x = - \left(\frac{3}{2}\right)$

Or,

$x - 2 = 0$

or, $x = 2$

:.The solution for $x = 2 , - \frac{3}{2}$.