How do you write #y = -3/2x + 3# in standard form?

1 Answer
Jun 14, 2017

#3x+2y=6#

Explanation:

Standard Form looks like :
#ax+by=c#
where #a,b# and #c# are constants.

So we need to rearrange the formula to get both #x# and #y# to one side, and a number on the other side.

So from:

#y=-3/2x+3#

We should get rid of the fraction first. To do this, multiply everything by 2:

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

#=>2y=(2*-3)/2x+6#

#=>2y=(cancel(2)*-3)/cancel(2)x+6#

#=>2y=-3x+6#

Then we bring -3x to the other side by adding 3x to both sides:

#=>3x+2y=-3x+6+3x#

#=>3x+2y=cancel(-3x)+6+cancel(3x)#

#=>3x+2y=6#

And you've got it in standard from now