### Video Transcript

In the table below, some of the π₯-
and π¦-values are missing from the ordered pairs that satisfy the linear relation
negative five π₯ plus π¦ is equal to negative three. Find the missing π₯- and
π¦-values.

We can find the missing value in
each ordered pair by substituting the known π₯- or π¦-value into the equation for
the linear relation. Thatβs into negative five π₯ plus
π¦ is equal to negative three. We then solve for the missing π₯-
or π¦-value. Alternatively, we can rearrange the
equation, making the missing value the subject, and then solve.

For our first ordered pair, since
we want to solve for π¦, letβs rearrange the equation so that π¦ is the subject. We can do this by adding five π₯ to
both sides, which gives us π¦ is equal to negative three plus five π₯. And now substituting π₯ is equal to
negative four, this gives us π¦ is equal to negative three plus five times negative
four. That is, π¦ is negative three minus
20, which is negative 23. Our first ordered pair is therefore
negative four, negative 23.

Now, in our second ordered pair, we
know the value of π¦. That is, π¦ is equal to negative
13. So, since we want to find the value
of π₯, letβs make π₯ the subject of our relation. Subtracting π¦ from both sides of
negative five π₯ plus π¦ is negative three gives us negative five π₯ is equal to
negative three minus π¦. And now dividing through by
negative five, we have π₯ is equal to three over five plus π¦ over five. And we can rewrite this as π₯ is
equal to three plus π¦ over five. And now substituting π¦ is equal to
negative 13 into this equation, we have π₯ is equal to three plus negative 13
divided by five, that is, negative 10 over five, which is equal to negative two. Hence, the π₯-value in our second
ordered pair is negative two. And so our ordered pair is negative
two, negative 13.

Now, for our third ordered pair, we
know that π₯ is equal to zero. And into our equation for π¦, this
gives us π¦ is equal to negative three plus five times zero. That is, π¦ is equal to negative
three. Our third ordered pair is therefore
zero, negative three.

Since we again have the π₯-value β
thatβs π₯ is equal to three β for our fourth ordered pair, we can use the equation
for π¦ again. And so we have π¦ is equal to
negative three plus five times three, that is, negative three plus 15, which is
12. Our fourth ordered pair is
therefore three, 12.

And now for our fifth and final
ordered pair, weβre given that π¦ has the value of 32. And substituting this into our
equation for π₯, this gives us π₯ is equal to three plus 32 over five, that is, 35
divided by five, which is seven. And our final ordered pair is
therefore seven, 32.

The missing values from the table
are therefore π¦ is equal to negative 23, π₯ is negative two, π¦ is negative three,
π¦ is 12, and π₯ is equal to seven. And hence, our ordered pairs are
negative four, negative 23; negative two, negative 13; zero, negative three; three,
12; and seven, 32.