How do you solve # 2x – y = 4#, #3x + y = 1# by graphing and classify the system?

1 Answer
Aug 19, 2017

Answer:

See a solution process below:

Explanation:

For each of the equations we can find two points on the line and then draw a line through the points to graph the line.

Equation 1

For #x = 0#:

#(2 * 0) - y = 4#

#0 - y = 4#

#-y = 4#

#color(red)(-1) xx -y = color(red)(-1) xx 4#

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

For #x = 2#

#(2 * 2) - y = 4#

#4 - y = 4#

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

#0 - y = 0#

#-y = 0#

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

#y = 0# or #(2, 0)#

graph{(x^2+(y+4)^2-0.05)((x-2)^2+y^2-0.05)(2x-y-4)=0 [-15, 15, -7.5, 7.5]}

Equation 2

For #x = 0#:

#(3 * 0) + y = 1#

#0 + y = 1#

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

For #x = 2#

#(3 * 2) + y = 1#

#6 + y = 1#

#-color(red)(6) + 6 + y = -color(red)(6) + 1#

#0 + y = -5#

#y = -5# or #(2, -5)#

graph{(3x+y-1)(x^2+(y-1)^2-0.05)((x-2)^2+(y+5)^2-0.05)(2x-y-4)=0 [-15, 15, -7.5, 7.5]}

Solution

We can see the lines cross at #(1, -2)#

Therefore:

The system is consistent because it has at least one solution. And, it is independent because the lines have different slopes.

graph{(3x+y-1)((x-1)^2+(y+2)^2-0.015)(2x-y-4)=0 [-6, 6, -3, 3]}