Question

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`?

Answer 1

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.