# How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent 12x + 4y = -4 and 2x -y =6?

Aug 25, 2017

See a solution process below:

#### Explanation:

To graph each line we need to find two points of solution for the equation, map the points then draw a line through the two points:

Equation 1

For $x = - 2$

$\left(12 \times - 2\right) + 4 y = - 4$

$- 24 + 4 y = - 4$

$\textcolor{red}{24} - 24 + 4 y = \textcolor{red}{24} - 4$

$0 + 4 y = 20$

$4 y = 20$

$\frac{4 y}{\textcolor{red}{4}} = \frac{20}{\textcolor{red}{4}}$

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

For $x = 0$

$\left(12 \times 0\right) + 4 y = - 4$

$0 + 4 y = - 4$

$4 y = - 4$

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

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

graph{((x+2)^2+(y-5)^2-0.05)(x^2+(y+1)^2-0.05)(12x+4y+4)=0 [-15, 15, -7.5, 7.5]}

Equation 2

For $x = 0$

$\left(2 \times 0\right) - y = 6$

$0 - y = 6$

$- y = 6$

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

$y = - 6$ or $\left(0 , - 6\right)$

For $y = 0$

$2 x - 0 = 6$

$2 x = 6$

$\frac{2 x}{\textcolor{red}{2}} = \frac{6}{\textcolor{red}{2}}$

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

graph{(2x-y-6)((x-3)^2+y^2-0.05)(x^2+(y+6)^2-0.05)(12x+4y+4)=0 [-15, 15, -7.5, 7.5]}

We can see the lines intersect at $\left(1 , - 4\right)$ for the solution to the problem.

graph{(2x-y-6)((x-1)^2+(y+4)^2-0.0125)(12x+4y+4)=0 [-8, 8, -7, 1]}

Because there is at least one solution the system of equations is consistent.