How do you solve the system #5x-y=4# and #-2x+6y=4# by graphing?

1 Answer
Aug 8, 2017

Answer:

See a solution process below:

Explanation:

First, solve for two points on the first line, plot the points and draw a line through the points:

For #x = 0#: #(5 * 0) - y = 4#

#0 - y = 4#

#-y = 4#

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

For #x = 2#: #(5 * 2) - y = 4#

#10 - y = 4#

#-y = -6#

#y = 6# or #(2, 6)#

graph{(5x - y - 4)(x^2 + (y + 4)^2 - 0.125)((x-2)^2 + (y - 6)^2 - 0.125) = 0 [-20, 20, -10, 10]]}

Do the same for the second equation:

For #x = 0#: #(-2 * 0) + 6y = 4#

#0 + 6y = 4#

#6y = 4#

#y = 2/3#

For #x = 2#: #(-2 * 2) + 6y = 4#

#-4 + 6y = 4#

#6y = 8#

#y = 4/3#

graph{(-2x + 6y - 4)(x^2+(y - (2/3))^2-0.125)((x-2)^2+(y - (4/3))^2-0.125) (5x - y - 4)(x^2 + (y + 4)^2 - 0.125)((x-2)^2 + (y - 6)^2 - 0.125) = 0 [-20, 20, -10, 10]]}

Zooming in, we can see the lines intersect at: #(1, 1)#

graph{(-2x + 6y - 4)(5x - y - 4)((x -1)^2+(y-1)^2-0.0125) = 0 [-6, 6, -3, 3]}