### Video Transcript

Answer the following questions for
the functions 𝑦 equals three 𝑥 minus one and 𝑦 equals negative a half 𝑥 plus
five over two. Complete the table of values for 𝑦
equals three 𝑥 minus one. Then, complete the table of values
for 𝑦 equals negative a half 𝑥 plus five over two.

There’s also a third part of the
question that we’ll come on to later. So to solve this problem, what
we’re gonna do first of all is find out what our 𝑦-values are for the given
𝑥-values and complete our tables. So we’re gonna start with 𝑦 equals
three 𝑥 minus one. So to find out what this is gonna
be for an 𝑥-value of negative two, we’re gonna substitute this in instead of
𝑥. So when we do this, we’re gonna get
𝑦 equals three multiplied by negative two minus one. And that’s because we substituted
in 𝑥 equals negative two.

So then next, what we’re gonna do
is have 𝑦 equals negative six minus one. And that’s because three multiplied
by negative two is negative six. So therefore, we get a 𝑦-value of
negative seven. So therefore, we can say that when
𝑥 equals negative two, 𝑦 will equal negative seven. So what we’re gonna do now is move
on to the next value of 𝑥, which is negative one. So we substitute this into our
equation. So when we do that, we’re gonna get
𝑦 equals three multiplied by negative one minus one, which is gonna give us 𝑦
equals negative three minus one. Which is gonna give us a value of
𝑦 of negative four. So we know that when 𝑥 equals
negative one, 𝑦 equals negative four.

So now, let’s move on to the next
value. So when 𝑥 equals zero, we’re gonna
have 𝑦 equals three multiplied by zero minus one. So this is gonna give us 𝑦 equals
zero minus one. So then, what we get is 𝑦 is equal
to negative one. So we can say that when 𝑥 is equal
to zero, 𝑦 is equal to negative one. Well, now, before we substitute in
our next value, it’s worth looking at a little pattern we can notice. We’ve gone from negative seven to
negative four, which means we’ve added three. Then from negative four to negative
one, so we’ve added three again. So we’ve shown that as the
𝑥-values increase by one, then what’s gonna happen to the 𝑦-values is that these
increase by three.

And this is in fact correct cause
if we take a look at 𝑦 equals three 𝑥 minus one, then what we’ve got here is the
coefficient of 𝑥 is three or positive three. So therefore, it’s this that tells
us what’s the difference between our 𝑦-terms are. So every time we move up an
𝑥-value of one, then our 𝑦-value is gonna increase by positive three. So therefore, our prediction for
the next value is that if 𝑥 is equal to one, 𝑦 is equal to two. Well, let’s try this out. Well, if we substitute in 𝑥 equals
one into our equation, we’re gonna get 𝑦 equals three multiplied by one minus one,
which is gonna give us 𝑦 equals three minus one. So we get the value we’re
expecting, which is 𝑦 is equal to two.

So, it’s worth noting, though, at
this point that this method really only works if our 𝑥-values are going up by one
each time. It could work if there was a
different change between our 𝑥-values, but we’d have to look more carefully at the
gap between them and how we’d work that out. So now, we’ve seen that our pattern
works, then all we need to do is add three to our 𝑦-value which we’ve just worked
out, which was two. And when we do that, we get
five. So therefore, we can see that our
table is now completed because the last value is that when 𝑥 is equal to two, 𝑦 is
equal to five. So now, we completed our first
table. Let’s move on to the next table of
values.

So now, for the next table of
values, what we’re gonna do is substitute in, first of all, the value for 𝑥 of
negative two into our equation, which is 𝑦 equals negative a half 𝑥 plus five over
two. And when we do that, what we’re
gonna get is 𝑦 equals negative a half multiplied by negative two plus five over
two. So when we do this, what we’re
gonna get is 𝑦 equals one plus five over two. And we get that because negative a
half multiplied by negative two is a negative multiplied by negative, which gives us
a positive, so that’s positive one, and then plus five over two. So therefore, this is gonna give us
our first 𝑦-value of seven over two.

Okay, great! So now, what’s our next value? Well, what we could do is
substitute in negative one into our equation to find out what it’s going to be. However, we’ve already discovered a
method from the previous table which would be a quicker and easier way to solve our
problem. And that’s because if we look,
we’ve got 𝑥-values of negative two, negative one, zero, one, and two. So they go up by one each time. So therefore, what we can do is
just add on the coefficient of our 𝑥-term in order to find out what each value
is.

Well, with this equation, we could
see that the coefficient of our 𝑥-term is negative a half. So we’re gonna add negative a half,
which is the same as subtracting a half. Well, if we subtract a half from
seven over two, well six over two is just three, so that’s our next 𝑦-value. And then if we subtract another
half, we get five over two. And then subtracting another half
gives us four over two, which is just the same as two. And then finally, we subtract our
final half away from two, which gives us three over two. So therefore, we completed the
second part of the question cause our second table is now complete.

So now, let’s take a look at the
third part of the question. So for the final part of this
question, what we’re asked is to use the table of values to determine the
intersection point of the two graphs. Well, what is the intersection
point? Well, the intersection point is the
point where the two graphs will meet or intersect. But in order to see if we can use
our tables to decide what the intersection point is going to be, what we’re looking
for is a pair of coordinates that are going to be the same in each table. And we can see that they’ve each
got this pair here. When 𝑥 equals one, 𝑦 equals
two. So therefore, we can say that the
intersection point is going to be the point one, two.