Question

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

Answer 1

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

Explanation:

Given: `x/(2x+1)=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 `2x+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 - 4x` to both sides:

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

Factor the quadratic:

`(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`

Check:

`-5/(2(-5)+1)=5/(4--5)`
`-1/(2(-1)+1)=5/(4--1)`

`5/9=5/9`
`1=1`

This checks