How do you graph #3y+4x=12#?

1 Answer
Feb 15, 2017

Answer:

This is a straight line through #(0, 4)# and #(3, 0)#

Explanation:

Notice that the terms are linear or constant, so this equation represents a straight line.

We can find the intersections with the #x# and #y# axes by setting #y# or #x# equal to zero, or equivalently covering up the corresponding term and solving the resultant equation.

So putting #x=0# we get:

#3y = 12#

Dividing both sides by #3#, we find:

#y=4#

So the line intercepts the #y# axis (which has equation #x=0#) at the point #(0, 4)#

If we instead put #y=0# then we get:

#4x=12#

and hence:

#x=3#

So the #x# intercept is at #(3, 0)#

We can now draw our line through these two intercepts:
graph{(4x+3y-12)((x-3)^2+y^2-0.01)(x^2+(y-4)^2-0.01) = 0 [-9.21, 10.79, -2.28, 7.72]}