# How do you solve [ x/(x+1)] = [1/3] + [(x+2)/(4x)]?

Feb 18, 2016

$x = 3$ or $x = - \frac{2}{5}$

#### Explanation:

first of all equalize all the denominators as if you're doing a standard fraction addition/subtraction

the right-hand side
x/(x+1)= (4x*1/(3*4x))+(3*(x+2))/(3*4x))

$\frac{x}{x + 1} = \frac{4 x + 3 x + 6}{12 x}$

$\frac{x}{x + 1} = \frac{7 x + 6}{12 x}$

$\frac{x}{x + 1} - \frac{7 x + 6}{12 x} = 0$

Now the left-hand side

$\left(\frac{x \cdot 12 x}{12 x \cdot \left(x + 1\right)}\right) - \frac{\left(x + 1\right) \cdot \left(7 x + 6\right)}{\left(x + 1\right) \cdot 12 x} = 0$

$\frac{12 {x}^{2} - \left(7 {x}^{2} + 6 x + 7 x + 6\right)}{12 x \cdot \left(x + 1\right)} = 0$

$\frac{12 {x}^{2} - \left(7 {x}^{2} + 13 x + 6\right)}{12 x \cdot \left(x + 1\right)} = 0$

$\frac{12 {x}^{2} - 7 {x}^{2} - 13 x - 6}{12 x \cdot \left(x + 1\right)} = 0$

$\frac{5 {x}^{2} - 13 x - 6}{12 x \cdot \left(x + 1\right)} = 0$

the nominator of this fractional equation should be 0 for the expression to yield 0.
So solve for x in:

$\left(5 {x}^{2} - 13 x - 6\right) = 0$

Use the quadratic formula (or factorize the expression)

$x = - \frac{2}{5}$ or $x = 3$

Feel free to ask questions if you have any