# How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent 2x - 3y = 14 and 2x + y = -10?

Aug 21, 2017

See a solution process below:

#### Explanation:

To graph the lines first find two points on the line. Then plot the points. Then draw a line through the points:

Equation 1:

For $x = 1$

$\left(2 \cdot 1\right) - 3 y = 14$

$2 - 3 y = 14$

$- 3 y = 12$

$y = - 4$ or $\left(1 , - 4\right)$

For $x = 4$

$\left(2 \cdot 4\right) - 3 y = 14$

$8 - 3 y = 14$

$- 3 y = 6$

$y = - 2$ or $\left(1 , - 2\right)$

graph{((x-1)^2+(y+4)^2-0.25)((x-4)^2+(y+2)^2-0.25)(2x-3y-14)=0 [-30, 30, -15, 15]}

Equation 2:

For $x = 0$

$\left(2 \cdot 0\right) + y = - 10$

$0 + y = - 10$

$y = - 10$ or $\left(0 , - 10\right)$

For $x = - 5$

$\left(2 \cdot - 5\right) + y = - 10$

$- 10 + y = - 10$

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

graph{(2x+y+10)(x^2+(y+10)^2-0.25)((x+5)^2+y^2-0.25)(2x-3y-14)=0 [-30, 30, -15, 15]}

We can see the the lines cross at $\left(- 2 , - 6\right)$ Because there is on solution the system is consistent.

graph{((x+2)^2+(y+6)^2-0.05)(2x+y+10)(2x-3y-14)=0 [-16, 16, -8, 8]}