How do you solve for y #9x + 3y - 5 = 6x - 9y + 10#?

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Answer 1 · 2015-10-11T00:00:52

#(5-x)/4#

Your goal here is to isolate #y# on one side of the equation.

This implies that you need to have all the #x#-terms and all the integers on the other side of the equation. Your starting equation looks like this

#9x + 3y - 5 = 6x - 9y + 10#

Start by adding #5# to both sides of the equation

#9x + 3y - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5))) = 6x - 9y + 10 + 5#

#9x + 3y = 6x - 9y + 15#

Next, add #9y# to both sides of the equation

#9x + 3y + 9y = 6x - color(red)(cancel(color(black)(9y))) + color(red)(cancel(color(black)(9y))) + 15#

#9x + 12y = 6x + 15#

Next, add #-9x# to both sides

#color(red)(cancel(color(black)(9x))) - color(red)(cancel(color(black)(9x))) + 12y = 6x - 9x + 15#

#12y = -3x + 15#

Finally, divide both sides by #12# and simplify where possible

#(color(red)(cancel(color(black)(12)))y)/color(red)(cancel(color(black)(12))) = (-3x + 15)/12#

#y = color(green)((5-x)/4)#