### Video Transcript

Given the linear relation negative
four π₯ plus two π¦ is equal to negative six, complete the table of values
below. Using the complete table, sketch a
graph to represent the relation.

In this example, we first need to
fill in the missing values in the table, where each ordered pair π₯ and π¦ must
satisfy the given linear relation. We can then use the values from the
completed table to sketch a graph of the relation.

So letβs begin by reminding
ourselves of what we mean by a linear relation. A relation of the form ππ₯ plus
ππ¦ equals π is a linear relation that can be represented by a set of ordered
pairs π₯, π¦, where each pair of values satisfies the given equation for specific
constants π, π, and π. For four out of the five ordered
pairs, π₯, π¦, in the table, weβre given either an π₯-value or a π¦-value. And so we must solve the equation,
negative four π₯ plus two π¦ equals negative six, to find the missing value in each
of these four ordered pairs.

So letβs begin with the first
ordered pair, where weβre given that π¦ equals negative seven. And we must find the π₯-value that
together with this π¦-value satisfies the given equation. To do this, we substitute π¦ equals
negative seven into the equation, which gives us negative four π₯ minus 14 is equal
to negative six. Now, adding 14 to both sides, we
have negative four π₯ is equal to eight. And dividing both sides by negative
four, we isolate π₯ on the left-hand side. And weβre left with negative two on
the right. So, if π¦ is negative seven, then
the value of π₯ satisfying the given linear relation must be negative two. And we can put this value in the
table. Although weβre not specifically
asked to, we can write the ordered pair negative two, negative seven as shown below
the table.

In the second ordered pair, weβre
given the π₯-value, that is, π₯ equal to zero. So, this time, we have to solve for
π¦. Substituting π₯ equal to zero into
the equation for the linear relation, we have negative four times zero plus two π¦
equals negative six. Since negative four times zero is
equal to zero, this leaves us with two π¦ equals negative six. We can isolate π¦ on the left by
dividing both sides by two. And so weβre left with π¦ equals
negative three. Hence, when π₯ is zero in the
relation negative four π₯ plus two π¦ equals negative six, π¦ must be equal to
negative three. And we can put this value into the
table with the ordered pair zero, negative three below.

In the next column of the table,
weβre given both values: π₯ equals one and π¦ equals negative one. So the ordered pair is one,
negative one. Now, moving on to the fourth
column, where weβre given that π¦ equals positive three, with this value substituted
into the equation, we have negative four π₯ plus six equals negative six. Subtracting six from both sides and
dividing through by negative four, we find when π¦ equals positive three, π₯ is also
positive three.

Finally, in the last column, weβre
given that π₯ is equal to five. Substituting this into the
equation, we have negative 20 plus two π¦ equals negative six. Adding 20 to both sides and
dividing through by two, we find π¦ is equal to seven. And the ordered pair is then five,
seven. Hence, given the linear relation
negative four π₯ plus two π¦ equals negative six, we find the missing values are
negative two, negative three, positive three, and seven.

Now, for the second part of this
question, using the values from the completed table, weβre going to sketch a graph
to represent the given relation. Note that weβve already written out
the ordered pairs that are the coordinates of points on the line representing this
relation. So all we need to do is plot the
points with these coordinates on our graph and draw a line through them. The first point has coordinates π₯
is negative two and π¦ is negative seven, which we can plot on our graph as
shown. The second has coordinates zero,
negative three; the third, one, negative one; and the fourth and fifth points have
coordinates three, three and five, seven, respectively.

Our final step is to draw a line
through these five points. And with this line, weβve sketched
a graph that represents the relation negative four π₯ plus two π¦ equals negative
six.