### Video Transcript

Consider the equation three π¦
equals six π₯ plus three over two. Rearrange the equation into the
form π¦ equals ππ₯ plus π. What are the slope and π¦-intercept
of the equation? Use the slope and intercept to
identify the correct graph of the equation.

Letβs begin this question by having
a look at the first part. Here, weβre asked to rearrange our
equation into the form π¦ equals ππ₯ plus π. We can remember that this form, π¦
equals ππ₯ plus π or often π¦ equals ππ₯ plus π, is the general form which
allows us to identify key parts of a graph. We can see that our equation has a
π¦ and an π₯ in the same way that the general form has a π¦ and an π₯. The difference here is that we have
a three π¦ instead of just π¦. So weβll need to divide both sides
of our equation by three.

This means that, on the right-hand
side, weβll be dividing by two multiplied by three. And as two times three is six, we
have π¦ equals six π₯ plus three over six. In order to simplify this fraction,
we can consider the right-hand side as six π₯ over six plus three over six. This is equivalent to π¦ equals π₯
plus a half. And this means we have rearranged
the equation into the form π¦ equals ππ₯ plus π. For the second part, we can recall
that when we have π¦ equals ππ₯ plus π, then the coefficient of π₯, the letter π,
indicates the slope or gradient of an equation. The constant term π represents the
π¦-intercept.

So when we take our equation in
this form, thatβs π¦ equals π₯ plus a half, the slope will be represented by the
coefficient of π₯, which in this case would be one. The π¦-intercept will be positive
one-half, which we can write as a half. And so we have answered the second
question. We can now take a look at the graph
options for the third part of the question. Weβve established that the
π¦-intercept of our graph will be at a half. We can see on our first graph that
the π¦-intercept is a half since the graph crosses through half on the π¦-axis. So this may be a possible
answer.

On the second graph, it crosses the
π¦-axis at negative a half. And so we can rule out the second
graph. The third graph has a π¦-intercept
of one, so this does not fit. And the fourth has a π¦-intercept
at negative one. The final graph has a π¦-intercept
of negative a half. So this wouldnβt work either. This means that we have one answer
left, but it is worth checking if the slope is equal to one.

We can recall that, to find the
slope of a line between two coordinates π₯ one, π¦ one and π₯ two, π¦ two, we
calculate π¦ two minus π¦ one over π₯ two minus π₯ one. We can select any two coordinates
that lie on the line. Here we have zero, a half and one,
three over two. It doesnβt matter which one we
designate as π₯ one, π¦ one and which we designate as π₯ two, π¦ two. So to find the slope, we have
three-halves subtract a half over one minus zero. And since three-halves subtract a
half is one and one subtract zero is one, we have one over one, which means that the
slope is indeed equal to one. And so our answer is that it is the
first graph which represents the equation three π¦ equals six π₯ plus three over
two.