# How do you solve  4x - y = 30,  4x + 5y = -6 by graphing and classify the system?

Jan 16, 2018

See a solution process below:

#### Explanation:

To solve this problem, plot two points for each equation and draw a line through the points.

Equation 1:

For $x = 0$

$\left(4 \cdot 0\right) - y = 30$

$0 - y = 30$

$- y = 30$

$- 1 \cdot - y = - 1 \cdot 30$

$y = - 30$ or $\left(0 , - 30\right)$

For $x = 5$

$\left(4 \cdot 5\right) - y = 30$

$20 - y = 30$

$20 - \textcolor{red}{20} - y = 30 - \textcolor{red}{20}$

$0 - y = 10$

$- y = 10$

$- 1 \cdot - y = - 1 \cdot 10$

$y = - 10$ or $\left(5 , - 10\right)$

graph{(x^2 + (y + 30)^2 - 0.75)((x - 5)^2 + (y + 10)^2 - 0.75)(4x - y - 30) = 0 [-100, 100, -50, 50]}

Equation 2:

For $x = 0$

$\left(4 \cdot 0\right) + 5 y = - 6$

$0 + 5 y = - 6$

$5 y = - 6$

$\frac{5 y}{\textcolor{red}{5}} = - \frac{6}{\textcolor{red}{5}}$

$y = - \frac{6}{5}$ or $\left(0 , - \frac{6}{5}\right)$

For $y = 0$

$4 x + \left(5 \cdot 0\right) = - 6$

$4 x + 0 = - 6$

$4 x = - 6$

$\frac{4 x}{\textcolor{red}{4}} = - \frac{6}{\textcolor{red}{4}}$

$x = - \frac{3}{2}$ or $\left(- \frac{3}{2} , 0\right)$

graph{(x^2 + (y + 6/5)^2 - 0.75)((x + 3/2)^2 + y^2 - 0.75)(4x + 5y + 6)(4x - y - 30) = 0 [-100, 100, -50, 50]}

We can see the lines cross at: $\left(6 , - 6\right)$

These two equations are consistent and independent because the have a solution and only one solution.

graph{((x-6)^2 + (y + 6)^2 - 0.05)(4x + 5y + 6)(4x - y - 30) = 0 [-2, 14, -7, 1]}