How do you solve #24x +2y = 48# for #y#?

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Answer 1 · 2017-09-09T01:57:10

See a solution process below:

First, subtract #color(red)(24x)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#-color(red)(24x) + 24x + 2y = -color(red)(24x) + 48#

#0 + 2y = -24x + 48#

#2y = -24x + 48#

Now, divide each side of the equation by #color(red)(2)# to solve for #y# while keeping the equation balanced:

#(2y)/color(red)(2) = (-24x + 48)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = (-24x)/color(red)(2) + 48/color(red)(2)#

#y = -12x + 24#