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

Aug 19, 2017

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 $x = 0$:

$\left(2 \cdot 0\right) - y = 4$

$0 - y = 4$

$- y = 4$

$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times 4$

$y = - 4$ or $\left(0 , - 4\right)$

For $x = 2$

$\left(2 \cdot 2\right) - y = 4$

$4 - y = 4$

$- \textcolor{red}{4} + 4 - y = - \textcolor{red}{4} + 4$

$0 - y = 0$

$- y = 0$

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

$y = 0$ or $\left(2 , 0\right)$

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 $x = 0$:

$\left(3 \cdot 0\right) + y = 1$

$0 + y = 1$

$y = 1$ or $\left(0 , 1\right)$

For $x = 2$

$\left(3 \cdot 2\right) + y = 1$

$6 + y = 1$

$- \textcolor{red}{6} + 6 + y = - \textcolor{red}{6} + 1$

$0 + y = - 5$

$y = - 5$ or $\left(2 , - 5\right)$

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 $\left(1 , - 2\right)$

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]}