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

Question

Nathan Axel · Accepted Answer

#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!

Abigail Allen · Answer

undefined To solve for $ y $ in the equation $ 2x + 3y = 3 $, you would first isolate the variable $ y $ by moving the term containing $ x $ to the other side of the equation. To do this, subtract $ 2x $ from both sides:

$$ 2x + 3y - 2x = 3 - 2x $$

This simplifies to:

$$ 3y = 3 - 2x $$

Next, divide both sides by 3 to solve for $ y $:

$$ \frac{3y}{3} = \frac{3 - 2x}{3} $$

This simplifies to:

$$ y = \frac{3 - 2x}{3} $$