How do you solve the system #4x + 6y = -12 # and #2x + 3y = 9# by graphing?

1 Answer

Answer:

See below:

Explanation:

I find that the way I like to graph lines is to put the equations into slope-intercept form. The general formula is:

#y=mx+b#

where #m# is the slope and #b# is the #y#-intercept. I like this form because I'm given a point (the #y#-intercept) and a straightforward way to generate a second point (which is all I need to then pull out a straightedge and connect the points).

Let's convert the two equations:

#4x+6y=-12#
#2x+3y=9#

#6y=-4x-12#
#3y=-2x+9#

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

Before we graph these points, notice that the slopes are the same - this means the two lines will be either parallel or the same line. Also notice that the 2 #y#-intercepts are different, and so we're not dealing with the same line written two different ways but are instead dealing with two parallel lines.

To graph these lines, first we mark the #y#-intercepts (#-2# for the first equation and #3# for the second).

To find our second points, remember that slope can be found by using the "formula" #"rise"/"run"#. The numerator tells us the number of points to rise (or fall - with negative slope I like to handle it in the numerator) against the number of points to run (i.e. move to the right) shown in the denominator. For instance, with the first line, we'd start at #(0,-2)# and then move two points down and three points right, giving:

#(0+3, -2-2)=(3,-4)#

Doing the same for the other line gives

#(0+3,3-2)=(3,1)#

The graph itself looks like this:

graph{(4x+6y+12)(2x+3y-9)((x-0)^2+(y+2)^2-.1)((x-0)^2+(y-3)^2-.1)((x-3)^2+(y+4)^2-.1)((x-3)^2+(y-1)^2-.1)=0}