Video Transcript
In this video, we’ll learn how to
identify and graph linear relations between two variables given the relation in the
form 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐 and write the ordered pairs that satisfy the
given equation.
So let’s begin by defining what we
mean by a linear relation. If two variables 𝑥 and 𝑦 are
related by an equation of the form 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐, where 𝑎, 𝑏, and
𝑐 are real numbers, then 𝑥 and 𝑦 are linearly related. Such a relation can be represented
by a set of ordered pairs 𝑥, 𝑦. So let’s look at an example.
Suppose the cost of sending a
birthday card in the post is 2.4 Egyptian pounds. If only 20-piaster and 50-piaster
stamps are available, write the ordered pairs for how many of each we could use to
send a birthday card. And we note that 100 piaster is
equal to one Egyptian pound.
Since 2.4 Egyptian pounds is equal
to 240 piaster, the total value of the stamps must be 240 piaster. If we call 𝑥 the number of
20-piaster stamps and 𝑦 the number of 50-piaster stamps, then to represent the
postage, we have the linear relation 20𝑥 plus 50𝑦 is equal to 240. And we want to know which, if any,
pairs of values of 𝑥 and 𝑦 together satisfy this equation.
Note that we cannot break these
stamps up into smaller values. So 𝑥 and 𝑦 must be natural
numbers, that is, positive whole numbers. The first ordered pair that
satisfies this equation is 𝑥 is equal to two and 𝑦 is equal to four. This gives us the ordered pair two,
four such that 20 times two plus 50 times four is equal to 240. That is, 40 piaster plus 200
piaster is equal to 240 piaster. Our second ordered pair is 𝑥 is
seven and 𝑦 is equal to two. This gives us the ordered pair
seven, two. This gives us 20 times seven plus
50 times two, that is, 140 piaster plus 100 piaster, which is equal to 240
piaster. Our final possible ordered pair is
𝑥 is 12 and 𝑦 is zero. This gives us 20 times 12 plus 50
times zero, which is 240 plus zero. Then, that’s 240 piaster. Our ordered pairs are then two,
four; seven, two; and 12, zero.
Note that each of the terms in our
linear relation has a common factor of 10. And dividing through our equation
by 10 gives us two 𝑥 plus five 𝑦 is equal to 24. And in fact each of our ordered
pairs also satisfies this new equation. And since there are no factors
common to each of the terms in this new equation, we call this the simplest form of
the linear relation. Let’s look now at an example of how
we determine whether or not an ordered pair satisfies a particular linear
relation.
Which of the following satisfies
the relation 𝑥 minus 𝑦 is equal to negative 10? Is it (A) the ordered pair negative
two, negative two? (B) The ordered pair negative 16,
six. (C) Nine, negative one. (D) Negative 12, negative two. Or (E) the ordered pair negative
five, negative 15.
To answer this question, we try
each pair of values in turn in the given equation. That’s 𝑥 minus 𝑦 is equal to
negative 10. This means that from each pair, we
substitute the first value for 𝑥 and the second for 𝑦. And if the right- and left-hand
sides are equal, then we can say that the pair satisfies the relation 𝑥 minus 𝑦 is
equal to negative 10.
So, looking first at option (A), we
have 𝑥 is equal to negative two and 𝑦 is also equal to negative two. Substituting these values into our
linear relation, this gives us negative two minus negative two. And we want to know, does this
equal negative 10? Now, since subtracting negative two
is the same as adding positive two, this gives us negative two plus two. But that’s equal to zero, which of
course is not equal to negative 10. This means that our first ordered
pair (A) does not satisfy the linear relation 𝑥 minus 𝑦 is equal to negative
10.
Next, we consider option (B). That’s where 𝑥 is negative 16 and
𝑦 is equal to six. Into our linear relation, this
gives us negative 16 minus six. And this evaluates to negative 22,
and this does not equal negative 10. And hence, we can say that option
(B) does not satisfy the linear relation.
For option (C), our values are 𝑥
is nine and 𝑦 is negative one. This gives us nine minus negative
one, which evaluates to 10, which is not equal to negative 10. Hence, option (C) does not satisfy
the linear relation.
For option (D), our values are 𝑥
is equal to negative 12 and 𝑦 is negative two. Into our linear relation 𝑥 minus
𝑦 is equal to negative 10, this gives us negative 12 minus negative two. And since subtracting negative two
is the same as adding positive two, we have negative 12 plus two, and that’s equal
to negative 10. And hence, option (D) does satisfy
the linear relation.
Finally, checking our option (E),
this has values 𝑥 is equal to negative five and 𝑦 is negative 15, which in our
linear relation give us negative five minus negative 15. And this evaluates to negative five
plus 15, which is 10. Since this is not equal to negative
10, we can say that option (E) does not satisfy the linear relation 𝑥 minus 𝑦 is
equal to negative 10.
We can see that only one of the
given ordered pairs satisfies the relation 𝑥 minus 𝑦 is equal to negative 10, that
is, option (D). Hence, the ordered pair negative
12, negative two satisfies the relation.
In our next example, we can see how
to find the missing value in an ordered pair given a linear relation.
In the table below, some of the 𝑥-
and 𝑦-values are missing from the ordered pairs that satisfy the linear relation
negative five 𝑥 plus 𝑦 is equal to negative three. Find the missing 𝑥- and
𝑦-values.
We can find the missing value in
each ordered pair by substituting the known 𝑥- or 𝑦-value into the equation for
the linear relation. That’s into negative five 𝑥 plus
𝑦 is equal to negative three. We then solve for the missing 𝑥-
or 𝑦-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 𝑦, let’s rearrange the equation so that 𝑦 is the subject. We can do this by adding five 𝑥 to
both sides, which gives us 𝑦 is equal to negative three plus five 𝑥. And now substituting 𝑥 is equal to
negative four, this gives us 𝑦 is equal to negative three plus five times negative
four. That is, 𝑦 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 𝑦. That is, 𝑦 is equal to negative
13. So, since we want to find the value
of 𝑥, let’s make 𝑥 the subject of our relation. Subtracting 𝑦 from both sides of
negative five 𝑥 plus 𝑦 is negative three gives us negative five 𝑥 is equal to
negative three minus 𝑦. And now dividing through by
negative five, we have 𝑥 is equal to three over five plus 𝑦 over five. And we can rewrite this as 𝑥 is
equal to three plus 𝑦 over five. And now substituting 𝑦 is equal to
negative 13 into this equation, we have 𝑥 is equal to three plus negative 13
divided by five, that is, negative 10 over five, which is equal to negative two. Hence, the 𝑥-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 𝑥 is equal to zero. And into our equation for 𝑦, this
gives us 𝑦 is equal to negative three plus five times zero. That is, 𝑦 is equal to negative
three. Our third ordered pair is therefore
zero, negative three.
Since we again have the 𝑥-value —
that’s 𝑥 is equal to three — for our fourth ordered pair, we can use the equation
for 𝑦 again. And so we have 𝑦 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, we’re given that 𝑦 has the value of 32. And substituting this into our
equation for 𝑥, this gives us 𝑥 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 𝑦 is equal to negative 23, 𝑥 is negative two, 𝑦 is negative three,
𝑦 is 12, and 𝑥 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.
In our next example, we’re going to
use a given linear equation and an ordered pair that satisfies that relation to find
an unknown coefficient.
Given that the ordered pair
negative seven, negative three satisfies the relation three 𝑥 plus 𝑏𝑦 is equal to
negative three, find the value of 𝑏.
To find the value of 𝑏 in the
linear relation given, we substitute the 𝑥- and 𝑦-values from the ordered pair
into the linear relation. We then solve the linear relation
for 𝑏. With 𝑥 is equal to negative seven
and 𝑦 is negative three then, into the linear relation three 𝑥 plus 𝑏𝑦 is
negative three, this gives us three times negative seven plus 𝑏 times negative
three is equal to negative three. That is, negative 21 minus three 𝑏
is equal to negative three. To solve for 𝑏, we can then add 21
to both sides, and this gives negative three 𝑏 is equal to negative three plus 21,
which evaluates on the right-hand side to 18.
And now dividing both sides by
negative three, we have 𝑏 is equal to 18 over negative three. That is, 𝑏 is equal to negative
six. Hence, the value of 𝑏 that
satisfies the relation three 𝑥 plus 𝑏𝑦 is equal to negative three given that the
point negative seven, negative three satisfies this relation is 𝑏 is negative
six.
A linear relation can be
represented graphically as a straight line, hence the term “linear.” If we know at least two ordered
pairs that satisfy a specific linear relation, to represent the relation
graphically, we plot the points represented by the ordered pairs and draw the line
that passes through both points. For example, the ordered pairs
negative one, negative three and two, three both satisfy the linear relation
negative two 𝑥 plus 𝑦 is equal to negative one. We can represent this relation
graphically by plotting the points with coordinates 𝑥 is equal to negative one, 𝑦
is negative three and 𝑥 is two, 𝑦 is three and drawing a line through those
points.
Note that while we’ve plotted the
line corresponding to the linear relation using only the two points given, in fact,
every point on the line is represented by an ordered pair 𝑥, 𝑦 that satisfies the
linear relation negative two 𝑥 plus 𝑦 is equal to negative one. And that’s where 𝑥 and 𝑦 are the
coordinates of the point on the line.
With this in mind, looking once
again at our postage stamps example, we note that the linear relation 20𝑥 plus 50𝑦
is equal to 240, which we can write equivalently as two 𝑥 plus five 𝑦 is 24, can
be represented by the graph shown. In this case, however, while the
linear relation 20𝑥 plus 50𝑦 is equal to 240 is represented completely by the
plotted line, we know that, in fact, there are only three points on that line that
correspond to the actual scenario of buying stamps to the value of 240 piaster. Remember that 𝑥 was the number of
20-piaster stamps and 𝑦 the number of 50-piaster stamps and these cannot be broken
down into smaller units. And so for this particular
scenario, the solutions must be positive whole-number values. While this is true for the number
of stamps, however, any point 𝑥, 𝑦 on the given line satisfies the linear relation
20𝑥 plus 50𝑦 is 240, or equivalently two 𝑥 plus five 𝑦 is equal to 24.
Now, in the linear relations
discussed so far, that is, relations of the form 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐, the
coefficients 𝑎 and 𝑏 have been nonzero. Let’s look now at the special cases
where either 𝑎 or 𝑏 is equal to zero.
In the linear relation 𝑎𝑥 plus
𝑏𝑦 is equal to 𝑐, in the special case where 𝑎 is equal to zero, the relation
reduces to 𝑏𝑦 is equal to 𝑐. If we then divide through by 𝑏, we
have 𝑦 is equal to 𝑐 over 𝑏. This means that for every value of
𝑥, 𝑦 is equal to the constant 𝑐 over 𝑏. What this means graphically is that
this is a horizontal line through the point on the 𝑦-axis where 𝑦 is equal to 𝑐
over 𝑏. So this is the case when 𝑎 is
equal to zero.
But if we now take the case where
𝑎 is nonzero and 𝑏 is equal to zero, we then have 𝑎𝑥 is equal to 𝑐. And now if we divide both sides by
𝑎, this gives us 𝑥 is equal to 𝑐 over 𝑎. And graphically, this is a vertical
line through the point on the 𝑥-axis where 𝑥 is equal to 𝑐 over 𝑎. In our next example, we’re asked to
plot the graph of a linear relation of this type.
Given the relation 𝑎𝑥 plus 𝑏𝑦
is equal to 𝑐, sketch its graph if 𝑎 is equal to zero, 𝑏 is three, and 𝑐 is
equal to two.
To sketch the graph of 𝑎𝑥 plus
𝑏𝑦 is equal to 𝑐, where the coefficients take the values given, we first
substitute these values into the equation. So, with 𝑎 equal to zero, 𝑏 equal
to three, and 𝑐 equal to two, we have zero times 𝑥 plus three times 𝑦 is equal to
two. That is, three 𝑦 is equal to
two. And dividing through by three, this
gives us 𝑦 is equal to two over three.
Before sketching this, let’s note
that another way of looking at this is that in the linear relation 𝑎𝑥 plus 𝑏𝑦 is
equal to 𝑐, if 𝑎 is equal to zero, the relation becomes 𝑦 is equal to 𝑐 over
𝑏. In the given problem, we have 𝑐
equal to two and 𝑏 equal to three. And as we found, 𝑦 is equal to two
over three. To sketch the graph of this
relation 𝑦 is equal to two over three, we note that the interpretation of this
relation is that for every value of 𝑥, 𝑦 is equal to the constant value two over
three. The graph of this relation then is
a horizontal line through the point on the 𝑦-axis where 𝑦 is equal to two over
three.
Let’s complete this video by
recapping the key points covered. A linear relation of the form 𝑎𝑥
plus 𝑏𝑦 is equal to 𝑐 can be represented by a set of ordered pairs 𝑥, 𝑦, where
each set of values satisfies the relation. Given an ordered pair 𝑥, 𝑦, we
may find an unknown coefficient in the relation 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐. To sketch the graph of a linear
relation, we plot the coordinates of two or more ordered pairs. The unique line through these
points represents the linear relation. Linear relations 𝑚𝑎𝑥 plus 𝑚𝑏𝑦
is equal to 𝑚𝑐 and 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐 are equivalent and have the same
graph. If 𝑎, 𝑏, and 𝑐 have no common
factors, then 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐 is the simplest form of the
relation. If 𝑎 is equal to zero in the
linear relation 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐, then 𝑦 is equal to 𝑐 over 𝑏, and
the graph is a horizontal line through 𝑦 is 𝑐 over 𝑏. And finally, if 𝑏 is equal to
zero, then 𝑥 is equal to 𝑐 over 𝑎, and the graph is a vertical line through 𝑥 is
𝑐 over 𝑎.
