# a) If x - 2y = -8, and 3x - 6y = -12, what are the values of x and y? b) How do you prove your answer from (a) graphically?

Mar 17, 2017

a). The first step in solving by substitution is always solving for one of the variables. Since x in the second equation has coefficient $1$, we'll choose this variable to isolate.

$x - 2 y = - 8 \to x = 2 y - 8$

We now substitute this into the first equation.

$3 \left(2 y - 8\right) - 6 y = - 12$

$6 y - 24 - 6 y = - 12$

$0 y = 12$

This is true for no real value of $y$, therefore this system has no real solution.

b). Let's do a little bit of work with the first equation.

$3 x - 6 y = - 12$

We factor out a $3$.

$3 \left(x - 2 y\right) = - 12$

Divide both sides by $3$

$x - 2 y = - 4$

We get an equation that is identical to the second on the left-hand side, but different on the right-hand side. What does this mean?

Suppose we were to graph both lines. We would first convert to slope-intercept form.

$x - 2 y = - 4 \to - 2 y = - 4 - x \to y = \frac{1}{2} x + 2$

For the second equation:

$x - 2 y = - 8 \to - 2 y = - 8 - x \to y = \frac{1}{2} x + 4$

These lines have equal slopes but different y-intercepts. This means that these are parallel lines, which is graphical proof that they will never intersect.

Hopefully this helps!