# 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?

Nov 23, 2017

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 $x = 0$

$\left(2 \cdot 0\right) - y = - 1$

$0 - y = - 1$

$- y = - 1$

$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times - 1$

$y = 1$ or (0, 1)

For $x = - 1$

$\left(2 \cdot - 1\right) - y = - 1$

$- 2 - y = - 1$

$\textcolor{red}{2} - 2 - y = \textcolor{red}{2} - 1$

$0 - y = 1$

$- y = 1$

$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times 1$

$y = - 1$ 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 $x = 0$:

$0 - y = 5$

$- y = 5$

$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times 5$

$y = - 5$ or $\left(0 , - 5\right)$

For $x = 5$:

$5 - y = 5$

$- \textcolor{red}{5} + 5 - y = - \textcolor{red}{5} + 5$

$0 - y = 0$

$- y = 0$

$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times 0$

$y = 0$ or $\left(5 , 0\right)$

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: $\left(- 6 , - 11\right)$

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.