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

1 Answer
Apr 3, 2017

Answer:

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