How do you graph #18x - 3y = 12#?

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Answer 1 · 2017-08-31T11:42:02

See a solution process below:

First, solve for two points which solve the equation and plot these points:

First Point:

For #x = 0#

#(18 * 0) - 3y = 12#

#0 - 3y = 12#

#-3y = 12#

#(-3y)/color(red)(-3) = 12/color(red)(-3)#

#y = -4# or #(0, -4)#

Second Point:

For #x = 1#

#(18 * 1) - 3y = 12#

#18 - 3y = 12#

#-color(red)(18) + 18 - 3y = -color(red)(18) + 12#

#0 - 3y = -6#

#-3y = -6#

#(-3y)/color(red)(-3) = -6/color(red)(-3)#

#y = 2# or #(1, 2)#

We can next graph the two points on the coordinate plane:

graph{(x^2+(y+4)^2-0.08)((x-1)^2+(y-2)^2-0.08)=0 [-20, 20, -10, 10]}

Now, we can draw a straight line through the two points to graph the line:

graph{(18x-3y-12)(x^2+(y+4)^2-0.08)((x-1)^2+(y-2)^2-0.08)=0 [-20, 20, -10, 10]}