0
-8x+3y=12 3y = 8x+12 y = (8/3)x + 4 So it will be a line with slope 8/3 and y intercept of 4
(3,3)
3y - 3 = 2y + 5 3y - 2y = 5 + 3 y = 8
-2x+3y=1 3y=1+2x y=(1+2x)/3 Then proceed to find points by plugging in given or arbitrary values of x.
Normally, this is a difficult form of the equation to deal with. To change it into a more familiar form, we can do this: 4x + 3y = 0 3y = -4x y = (-4/3)x
x = 3 and y = 0
2y + 3 = 3y + 9 2y - 3y = 9 - 3 or -y = 6 so y = -6
3y + 8 = 3 -8 -8 3y=-5 (3y)/3=-5/3 y=(-5/3)
The result will be a plane that intercepts the x-, y-, and z-axes at +9, +6, and +3, respectively.
y = -1
(-1, 0)
It is [4m-1, (4m-2)/3]