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

Question

Kennedy Adams · Accepted 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).

Given: #x/(2x+1)=5/(4-x)# Limiting the domain to prevent division by zero: #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# On both sides, apply the distributive property: #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# Divide the quadratic by: #(x + 5)(x + 1) = 0# Since it is obvious that x does not become either value, kindly take note that I have removed the restrictions. #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 verifies

Ethan Adkins · Answer

undefined To solve the equation x/(2x+1) = 5/(4-x), we can start by cross-multiplying to eliminate the fractions. This gives us x(4-x) = 5(2x+1). Expanding both sides of the equation, we get 4x - x^2 = 10x + 5. Rearranging the terms, we have -x^2 - 6x - 5 = 0. To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring the equation, we find (x+1)(x+5) = 0. Therefore, the solutions are x = -1 and x = -5.