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

Oct 15, 2017

See a solution process below:

#### Explanation:

For each of the equations solve the equation for two points:

Equation 1:

For $x = 0$:

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

$0 + y = 1$

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

For $x = 2$:

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

$4 + y = 1$

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

$0 + y = - 3$

$y = - 3$ or $\left(2 , - 3\right)$

Equation 2:

For $y = 0$:

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

$x + 0 = - 2$

$x = - 2$ or $\left(- 2 , 0\right)$

For $x = 0$:

$0 + 2 y = - 2$

$2 y = - 2$

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

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

Graph Equation 1 by plotting the two points and drawing a straight line through them:

graph{(x^2+(y-1)^2-0.03)((x-2)^2+(y+3)^2-0.03)(2x + y - 1)=0}

Add the graph Equation 2 by plotting the two points and drawing a straight line through them:

graph{(2x + y - 1)((x+2)^2+y^2-0.03)(x^2+(y+1)^2-0.03)(x+2y+2)=0}

Solution:

graph{(2x + y - 1)(x+2y+2)((x-1.3333333333)^2+(y+1.666666667)^2-0.0025)=0 [-2, 2, -2, 0]}

Or

$\left(\frac{8}{3} , - \frac{5}{3}\right)$