# How do you solve x/(2x+1)=5/(4-x)?

Apr 3, 2017

Restrict the domain to avoid division by 0.
Multiply both sides by each of the numerators.
Solve the resulting quadratic.

#### Explanation:

Given: $\frac{x}{2 x + 1} = \frac{5}{4 - x}$

Restricting the domain so that division by 0 is avoided:

x/(2x+1)=5/(4-x); x !=-1/2, x !=4

Multiply both sides of the equation by $2 x + 1$:

x=(5(2x+1))/(4-x); x !=-1/2, x !=4

Multiply both sides of the equation by $4 - x$:

x(4-x)=5(2x+1); x !=-1/2, x !=4

Use the distributive property on both sides:

4x-x^2=10x+5; x !=-1/2, x !=4

Add ${x}^{2} - 4 x$ to both sides:

0 =x^2+6x+5; x !=-1/2, x !=4

$\left(x + 5\right) \left(x + 1\right) = 0$

Please notice that I have dropped the restrictions, because it is clear that x does not become either value.

$x = - 5 \text{ and } x = - 1$

Check:

$- \frac{5}{2 \left(- 5\right) + 1} = \frac{5}{4 - - 5}$
$- \frac{1}{2 \left(- 1\right) + 1} = \frac{5}{4 - - 1}$

$\frac{5}{9} = \frac{5}{9}$
$1 = 1$

This checks