# How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent 6x - 7y = 49 and 7y - 6x = -49?

Jan 19, 2018

See a solution process below:

#### Explanation:

First, we need to graph each line by finding two solutions for each equation, plotting the solution points and then drawing a line through the points.

Equation 1:

• First Point: $x = 0$

$\left(6 \times 0\right) - 7 y = 49$

$0 - 7 y = 49$

$\frac{- 7 y}{\textcolor{red}{- 7}} = \frac{49}{\textcolor{red}{- 7}}$

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

• Second Point: $x = 7$

$\left(6 \times 7\right) - 7 y = 49$

$42 - 7 y = 49$

$42 - \textcolor{red}{42} - 7 y = 49 - \textcolor{red}{42}$

$0 - 7 y = 7$

$\frac{- 7 y}{\textcolor{red}{- 7}} = \frac{7}{\textcolor{red}{- 7}}$

$y = - 1$ or $\left(7 , - 1\right)$

-First Graph:

graph{(x^2 + (y+7)^2 - 0.075)((x - 7)^2 + (y+1)^2 - 0.075)(6x - 7y - 49) = 0 [-20, 20, -10, 10]}

Equation 2:

• First Point: $x = 0$

$7 y - \left(6 \times 0\right) = - 49$

$7 y - 0 = - 49$

$\frac{7 y}{\textcolor{red}{7}} = - \frac{49}{\textcolor{red}{7}}$

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

• Second Point: $x = 7$

$7 y - \left(6 \times 7\right) = - 49$

$7 y - 42 = - 49$

$7 y - 42 + \textcolor{red}{42} = - 49 + \textcolor{red}{42}$

$7 y - 0 = - 7$

$\frac{7 y}{\textcolor{red}{7}} = - \frac{7}{\textcolor{red}{7}}$

$y = - 1$ or $\left(7 , - 1\right)$

-First and Second Graph:

graph{(x^2 + (y+7)^2 - 0.075)((x - 7)^2 + (y+1)^2 - 0.075)(6x - 7y - 49) = 0 [-20, 20, -10, 10]}

Solutoin:

As we can see from the graph, both equations represent the same line. Therefore, there are an infinite number of solutions.

The lines are Consistent because there is at least one solution.