How do you solve the system #2x + 3y = 7# and #2x - 3y = 13# by graphing?

1 Answer
Feb 4, 2018

Answer:

See a solution process below:

Explanation:

To solve for a system of equations by graphing. Solve for two points we each equation, plot the points, draw a line through the points, determine where the two lines intersect:

Equation 1: #2x +3y = 7#

First Point: For #x = 2#

#(2 * 2) + 3y = 7#

#4 + 3y = 7#

#4 -color(red)(4) + 3y = 7 - color(red)(4)#

#0 + 3y = 3#

#3y = 3#

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

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

Second Point: For #y = 3#

#2x + (3 xx 3) = 7#

#2x + 9 = 7#

#2x + 9 - color(red)(9) = 7 - color(red)(9)#

#2x + 0 = -2#

#(2x)/color(red)(2) = -2/color(red)(2)#

#x = -1# or #(-1, 3)#

We can next plot the two points on the coordinate plane and draw a line through the two points:

graph{((x-2)^2+(y-1)^2-0.025)((x+1)^2+(y-3)^2-0.025)(2x+3y-7)=0 [-10, 10, -5, 5]}

Equation 2: #2x -3y = 13#

First Point: For #x = 2#

#(2 * 2) - 3y = 13#

#4 - 3y = 13#

#4 -color(red)(4) - 3y = 13 - color(red)(4)#

#0 - 3y = 9#

#-3y = 9#

#(-3y)/color(red)(-3) = 9/color(red)(-3)#

#y = -3# or #(2, -3)#

Second Point: For #y = -1#

#2x - (3 xx -1) = 13#

#2x + 3 = 13#

#2x + 3 - color(red)(3) = 13 - color(red)(3)#

#2x + 0 = 10#

#(2x)/color(red)(2) = 10/color(red)(2)#

#x = 5# or #(5, -1)#

We can next plot the two points on the coordinate plane and draw a line through the two points for the second equation:

graph{((x-2)^2+(y+3)^2-0.025)((x-5)^2+(y+1)^2-0.025)(2x+3y-7)(2x-3y-13)=0 [-10, 10, -5, 5]}

We can see the lines cross and #(5, -1)#

graph{((x-5)^2+(y+1)^2-0.05)(2x+3y-7)(2x-3y-13)=0 [-10, 10, -5, 5]}