How do you solve #u(u - 5) + 8u = u(u + 2) - 4 #?

Question

Julia Adams · Accepted Answer

See a solution process below:

Initially, multiply each term inside the parenthesis by the term outside the parenthesis to expand the terms in parenthesis on each side of the equation: #color(red)(u)(u - 5) + 8u = color(blue)(u)(u + 2) - 4# #(color(red)(u) xx u) - (color(red)(u) xx 5) + 8u = (color(blue)(u) xx u) + (color(blue)(u) xx 2) - 4# #u^2 - 5u + 8u = u^2 + 2u - 4# #u^2 + (-5 + 8)u = u^2 + 2u - 4# #u^2 + 3u = u^2 + 2u - 4# Next, subtract #color(red)(u^2)# from each side of the equation to eliminate this term while keeping the equation balanced: #-color(red)(u^2) + u^2 + 3u = -color(red)(u^2) + u^2 + 2u - 4# #0 + 3u = 0 + 2u - 4# #3u = 2u - 4# Now, subtract #color(red)(2u)# from each side of the equation to solve for #u# while keeping the equation balanced: #-color(red)(2u) + 3u = -color(red)(2u) + 2u - 4# #(-color(red)(2) + 3)u = 0 - 4# #1u = -4# #u = -4#

Hazel Anglin · Answer

undefined To solve the equation $ u(u - 5) + 8u = u(u + 2) - 4 $, first distribute and combine like terms:

$ u^2 - 5u + 8u = u^2 + 2u - 4 $

$ u^2 + 3u = u^2 + 2u - 4 $

Subtract $ u^2 $ from both sides:

$ 3u = 2u - 4 $

Subtract $ 2u $ from both sides:

$ u = -4 $