### Video Transcript

Use the graph shown to solve the
given simultaneous equations. π¦ equals negative two π₯ plus
three and π¦ equals three π₯ minus two.

When you solve by graphing, all you
need to do is look at the graph and decide where they intersect or where the lines
cross. So right away, we can actually see
that these two lines cross or intersect at the point one, one. From our graph, we had to go right
one up one. So π₯ is one and π¦ is one, so they
cross at the point one, one.

While this is our final answer,
letβs go ahead and take it a step further and decide which equation goes with which
line. Both of our equations are already
in slope intercept form, π¦ equals ππ₯ plus π, where π is your slope and π is
your π¦-intercept. So the slope is the rise over the
run. π, the π¦-intercept, is where you
crossed the π¦-axis.

So for our first equation, π¦
equals negative two π₯ plus three, we have a negative slope and a π¦-intercept of
three. So if we have a negative slope, if
you look at the line left or right, it should be going down; itβs decreasing. And with a π¦-intercept of three,
we should cross the π¦-axis at three. This would be the blue line.

Our other equation, π¦ equals three
π₯ minus two, three is our slope. So itβs positive. So our graph should be going up
left or right or increasing. And the π¦-intercept is negative
two, so we should be crossing the π¦-axis at negative two which does go with the
other line, the red line. So overall, we use the graph shown
to solve the given simultaneous equations by seeing where they intersected, and that
was at the point one, one.