How do you solve the system of equations #x+y=8#, #x-y=4# by graphing?

2 Answers
May 7, 2018

The solution is x=6, y=2: See explanation below.

Explanation:

To solve such a system you should regard each equation as a function of x and y, where
#x_1+y_1=8#, or #y_1=-x_1+8#
To be able to plot the graph you will note that x=0 gives y=8, and y=0 gives x=8, so the two points (0, 8) and (8, 0) are on the line.

#x_2-y_2=4#, or #y_2=x_2-4#
x=0 gives y=-4, and y=0 gives x= 4, so this line must go through the points (0, -4) and (4, 0)

This gives the following linear graphs:

enter image source here

The point you are interested in, has to be on both lines at the same time, i.e. it must be where the two lines cross. The graph shows that the two lines cross in (6,2), i.e. x=6, y=2, which, therefore, is the solution you want.

It is always good to check the solution one has got by incerting the found value in the two equations:
#x+y=8 => 6+2=8# Check
#x-y=4 => 6-2=4# Check

May 7, 2018

Plot the graphs of both equation and the point where the graph crosses each other is the answer. See the attached graph which shows the point of intersection (6,2). That means #x=6, y=2#.
enter image source here

Explanation:

First, we need to write both equations as #y=mx+b#

#x+y=8#
#(x+y)-x=8-x# ----> subtracting the #x# both sides
#y=8-x#
#y = -x+8# ----> equation 1

#x-y=4#
#(x-y)-x=4-x# ------> subtract #x# both sides
#-y=4-x#
#(-1)y=-1(4-x)#-----> multiply by #-1# both sides to make #y# the subject.
#y=-4+x#
#y=x-4#-----> equation 2

So we get:
#y = -x+8# ----> Red Graph
#y=x-4# ------> Blue Graph

Plot the graphs of both equation and the point where the graph crosses each other is the answer. See the attached graph which shows the point of intersection (6,2). That means #x=6, y=2#.

Check the answer:
#6 + 2 = 8#
#6-2=4#