How do you solve #(x+2)/(y-4)=3/4# for #y#?

1 Answer
Mar 18, 2018

See a solution process below:

Explanation:

First, because both sides of the equation are pure fractions we can flip the fractions and write the equation as:

#(y - 4)/(x + 2) = 4/3#

Next, multiply each side of the equation by #color(red)((x + 2))# to eliminate the fraction on the left side of the equation while keeping the equation balanced:

#color(red)((x + 2)) xx (y - 4)/(x + 2) = 4/3color(red)((x + 2))#

#cancel(color(red)((x + 2))) xx (y - 4)/color(red)(cancel(color(black)(x + 2))) = (4/3 xx color(red)(x)) + (4/3 xx color(red)(2))#

#y - 4 = 4/3x + 8/3#

Now, add #color(red)(4)# to each side of the equation to solve for #y# while keeping the equation balanced:

#y - 4 + color(red)(4) = 4/3x + 8/3 + color(red)(4)#

#y - 0 = 4/3x + 8/3 + (3/3 xx color(red)(4))#

#y = 4/3x + 8/3 + 12/3#

#y = 4/3x + 20/3#

Or

#y = 4/3(x + 5)#