How do you solve for y in #2x + 3y = 3#?

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Answer 1 · 2018-03-27T00:08:46

#y=-2/3x+1#

Move the #x# term to the right side, then divide everything by #3#:

#2x+3y=3#

#3y+2x=3#

#3y+2xcolor(blue)-color(blue)(2x)=3color(blue)-color(blue)(2x)#

#3ycolor(red)cancelcolor(black)(color(black)+2xcolor(blue)-color(blue)(2x))=3color(blue)-color(blue)(2x)#

#3y=3color(blue)-color(blue)(2x)#

#3y=-2x+3#

#color(blue)(color(black)(3y)/3)=color(blue)(color(black)(-2x+3)/3)#

#color(blue)(color(black)(color(red)cancelcolor(black)3y)/color(red)cancelcolor(blue)3)=color(blue)(color(black)(-2x+3)/3)#

#y=color(blue)(color(black)(-2x+3)/3)#

#y=color(blue)(color(black)(-2x)/3)+color(blue)(color(black)3/3)#

#y=color(blue)(color(black)(-2x)/3)+1#

#y=-2/3x+1#

That is solving for #y#. Hope this helped!