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

Jul 1, 2018

See a solution process below:

#### Explanation:

First, we need to plot two points for each line and then draw a line through the two points to graph each line:

EQUATION 1:

First Point: For $x = 0$

$\left(2 \cdot 0\right) + y = 3$

$0 + y = 3$

$y = 3$ or $\left(0 , 3\right)$

Second Point: For $y = 1$

$2 x + 1 = 3$

$2 x + 1 - \textcolor{red}{1} = 3 - \textcolor{red}{1}$

$2 x + 0 = 2$

$2 x = 2$

$\frac{2 x}{\textcolor{red}{2}} = \frac{2}{\textcolor{red}{2}}$

$x = 1$ or $\left(1 , 1\right)$

We can next plot the two points on the coordinate plane and draw a line through the two points:

graph{(2x + y - 3)(x^2+(y-3)^2-0.075)((x-1)^2+(y - 1)^2-0.075)=0 [-20, 20, -10, 10]}

EQUATION 2:

First Point: For $x = 0$

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

$0 + y = - 1$

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

Second Point: For $y = 1$

$2 x + 1 = - 1$

$2 x + 1 - \textcolor{red}{1} = - 1 - \textcolor{red}{1}$

$2 x + 0 = - 2$

$2 x = - 2$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{2}{\textcolor{red}{2}}$

$x = - 1$ or $\left(- 1 , 1\right)$

We can next plot the two points on the coordinate plane and draw a line through the two points:

graph{(2x + y - 3)(2x + y + 1)(x^2+(y+1)^2-0.075)((x+1)^2+(y - 1)^2-0.075)=0 [-20, 20, -10, 10]}

From the graphs of the lines it appears the lines are parallel so there are no points in common so there is no solution.

Or, the solution is the null or empty set: $\left\{\emptyset\right\}$