How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent x + 3y = 4  and 2x + 6y = 8?

Aug 24, 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 $x = - 2$

$- 2 + 3 y = 4$

$\textcolor{red}{2} - 2 + 3 y = \textcolor{red}{2} + 4$

$0 + 3 y = 6$

$3 y = 6$

$\frac{3 y}{\textcolor{red}{3}} = \frac{6}{\textcolor{red}{3}}$

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

For $x = 1$

$1 + 3 y = 4$

$- \textcolor{red}{1} + 1 + 3 y = - \textcolor{red}{1} + 4$

$0 + 3 y = 3$

$3 y = 3$

$\frac{3 y}{\textcolor{red}{3}} = \frac{3}{\textcolor{red}{3}}$

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

graph{((x+2)^2+(y-2)^2-0.025)((x-1)^2+(y-1)^2-0.025)(x+3y-4)=0}

Equation 2

For $x = - 2$

$\left(- 2 \times 2\right) + 6 y = 8$

$- 4 + 6 y = 8$

$\textcolor{red}{4} - 4 + 6 y = \textcolor{red}{4} + 8$

$0 + 6 y = 12$

$6 y = 12$

$\frac{6 y}{\textcolor{red}{6}} = \frac{12}{\textcolor{red}{6}}$

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

For $x = 1$

$\left(2 \times 1\right) + 6 y = 8$

$2 + 6 y = 8$

$- \textcolor{red}{2} + 2 + 6 y = - \textcolor{red}{2} + 8$

$0 + 6 y = 6$

$6 y = 6$

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

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

graph{((x+2)^2+(y-2)^2-0.025)((x-1)^2+(y-1)^2-0.025)(x+3y-4)=0}

As you can see both lines are graphed exactly the same.

Therefore there are an infinite number of solutions.

Then, the system is consistent because it has at least one common solution.

And, the system is dependent because it has an infinite number of solutions.