How do you solve $\frac{x}{2 x + 1} = \frac{5}{4 - x}$ $x/(2x+1)=5/(4-x)$ ?

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Answer 1 · 2017-04-03T16:43:17

Restrict the domain to avoid division by 0.
Multiply both sides by each of the numerators.
Solve the resulting quadratic.
Check your answer(s).

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

Restricting the domain so that division by 0 is avoided:

$\frac{x}{2 x + 1} = \frac{5}{4 - x}; x \neq - \frac{1}{2}, x \neq 4$ $x/(2x+1)=5/(4-x); x !=-1/2, x !=4$

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

$x = \frac{5 (2 x + 1)}{4 - x}; x \neq - \frac{1}{2}, x \neq 4$ $x=(5(2x+1))/(4-x); x !=-1/2, x !=4$

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

$x (4 - x) = 5 (2 x + 1); x \neq - \frac{1}{2}, x \neq 4$ $x(4-x)=5(2x+1); x !=-1/2, x !=4$

Use the distributive property on both sides:

$4 x - x^{2} = 10 x + 5; x \neq - \frac{1}{2}, x \neq 4$ $4x-x^2=10x+5; x !=-1/2, x !=4$

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

$0 = x^{2} + 6 x + 5; x \neq - \frac{1}{2}, x \neq 4$ $0 =x^2+6x+5; x !=-1/2, x !=4$

Factor the quadratic:

$(x + 5) (x + 1) = 0$ $(x + 5)(x + 1) = 0$

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

$x = - 5 and x = - 1$ $x = -5" and "x = -1$

Check:

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

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

This checks

How do you solve x/(2x+1)=5/(4-x)x2x+1=54−xx/(2x+1)=5/(4-x)?