How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent #2x-y=-1# and #x-y= 5#?

1 Answer
Nov 23, 2017

Answer:

See a solution process below:

Explanation:

We can graph each line by finding two points which solve the equation, plot the points and then drawing a straight line through them:

Equation 1:

For #x = 0#

#(2 * 0) - y = -1#

#0 - y = -1#

#-y = -1#

#color(red)(-1) xx -y = color(red)(-1) xx -1#

#y = 1# or #(0, 1)

For #x = -1#

#(2 * -1) - y = -1#

#-2 - y = -1#

#color(red)(2) - 2 - y = color(red)(2) - 1#

#0 - y = 1#

#-y = 1#

#color(red)(-1) xx -y = color(red)(-1) xx 1#

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

graph{(2x-y+1)(x^2+(y-1)^2-0.075)((x+1)^2+(y+1)^2-0.075)=0 [-10, 30, -15, 5]}

Equation 2:

For #x = 0#:

#0 - y = 5#

#-y = 5#

#color(red)(-1) xx -y = color(red)(-1) xx 5#

#y = -5# or #(0, -5)#

For #x = 5#:

#5 - y = 5#

#-color(red)(5) + 5 - y = -color(red)(5) + 5#

#0 - y = 0#

#-y = 0#

#color(red)(-1) xx -y = color(red)(-1) xx 0#

#y = 0# or #(5, 0)#

graph{(x - y - 5)(x^2+(y+5)^2-0.075)((x-5)^2+y^2-0.075)(2x-y+1)(x^2+(y-1)^2-0.075)((x+1)^2+(y+1)^2-0.075)=0 [-10, 30, -15, 5]}

From the graph, the solution is: #(-6, -11)#

graph{(x - y - 5)(2x-y+1)((x+6)^2+(y+11)^2-0.1)=0 [-10, 30, -15, 5]}

Because there is one solution which satisfies both equations, this system of equations is consistent.