# Question #cf6c6

Sep 6, 2017

See a solution process below:

#### Explanation:

Equation 1: Solve for two points which solve the first equation and plot these points:

First Point:

For $x = 0$

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

Second Point:

For $x = 2$

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

We can plot the two points on the coordinate plane and draw a line through them to graph the line:

graph{(y-x+2)(x^2+(y+2)^2-0.03)((x-2)^2+y^2-0.03)=0 [-15, 15, -7.5, 7.5]}

Equation 2: Solve for two points which solve the second equation and plot these points:

First Point:

For $x = 0$

$y = \left(- 2 \cdot 0\right) + 7$
$y = 0 + 7$
$y = 7$ or $\left(0 , 7\right)$

Second Point:

For $x = 2$

$y = \left(- 2 \cdot 2\right) + 7$
$y = - 4 + 7$
$y = 3$ or $\left(2 , 3\right)$

We can plot the two points on the coordinate plane and draw a line through them to graph the line:

graph{(y+2x-7)(x^2+(y-7)^2-0.05)((x-2)^2+(y-3)^2-0.05)(y-x+2)(x^2+(y+2)^2-0.05)((x-2)^2+y^2-0.05)=0 [-15, 15, -7.5, 7.5]}

Solution:

We can see from the graph, the lines cross at $\left(3. 1\right)$:

graph{((x-3)^2+(y-1)^2-0.05)(y+2x-7)(y-x+2)=0 [-10, 10, -2.5, 7.5]}