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Lesson Explainer: Linear Relations: π‘Žπ‘₯+𝑏𝑦=𝑐 Mathematics

In this explainer, we will learn how to identify and graph linear relations between two variables given the relation in the form π‘Žπ‘₯+𝑏𝑦=𝑐 and write the ordered pairs that satisfy the given equation.

Let’s begin by defining what we mean by a linear relation.

Definition: Linear Relations

If two variables π‘₯ and 𝑦 are related by an equation of the form π‘Žπ‘₯+𝑏𝑦=𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers, then π‘₯ and 𝑦 are linearly related. Such a relation can be represented by a set of ordered pairs (π‘₯,𝑦).

Suppose Sarah wants to send a birthday card to her cousin. It costs EGP 2.40 to send a card, and since Sarah has left it a little late in the day, the post office has only 20 pt and 50 pt stamps left (where pt stands for piastres and there are 100 piastres in one pound). Let’s consider how many 20 pt and 50 pt stamps she could use to send her card.

stamp 20 pt
stamp 50 pt

Since EGP 2.40 is equal to 240 pt, the total value of the stamps must be 240 pt. Let’s call the number of 20 pt stamps π‘₯ and the number of 50 pt stamps 𝑦. Then, to represent the postage, we have the linear relation 20π‘₯+50𝑦=240.

We want to know which, if any, pairs of values of π‘₯ and 𝑦 together satisfy this equation. Since we cannot break the stamps up into smaller values, π‘₯ and 𝑦 must be natural numbers, that is, positive whole numbers. In fact, the ordered pair (2,4) satisfies the equation. That is, π‘₯=2 and 𝑦=4, since (20Γ—2)+(50Γ—4)=40+200=240.

Alternatively, the ordered pair (7,2) also satisfies the linear relation, since 20Γ—7+50Γ—2=240. Hence, Sarah could either use 2Γ—20pt stamps and 4Γ—50pt stamps or 7Γ—20pt stamps and 2Γ—50pt stamps. In fact, there are 3 possible combinations of the two stamp values that satisfy the given linear relation, as shown in the table below.

(π‘₯)
20 pt Stamps
(𝑦)
50 pt Stamps
Ordered Pair
24(2,4)
72(7,2)
120(12,0)

Note that since each term in the equation of our linear relation has a common factor of 10, we can divide the equation through by 10. This gives 2π‘₯+5𝑦=24, and each of our ordered pairs also satisfies this new equation. Since there are now no factors that are common to all of the terms, 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.

Example 1: Determining Which Ordered Pairs Satisfy a Given Linear Relation

Which of the following satisfies the relation π‘₯βˆ’π‘¦=βˆ’10?

  1. (βˆ’2,βˆ’2)
  2. (βˆ’16,6)
  3. (9,βˆ’1)
  4. (βˆ’12,βˆ’2)
  5. (βˆ’5,βˆ’15)

Answer

To answer this question, we try each pair of values in turn in the given equation. To do this, from each pair, we substitute the first value for π‘₯ and the second for 𝑦, and if the right- and left-hand sides are equal, then we can say the pair satisfies the relation π‘₯βˆ’π‘¦=βˆ’10.

  1. (βˆ’2,βˆ’2)βˆΆβˆ’2βˆ’(βˆ’2)=4β‰ βˆ’10Γ—
  2. (βˆ’16,6)βˆΆβˆ’16βˆ’6=βˆ’22β‰ βˆ’10Γ—
  3. (9,βˆ’1)∢9βˆ’(βˆ’1)=10β‰ βˆ’10Γ—
  4. (βˆ’12,βˆ’2)βˆΆβˆ’12βˆ’(βˆ’2)=βˆ’10βœ“
  5. (βˆ’5,βˆ’15)βˆΆβˆ’5βˆ’(βˆ’15)=10β‰ βˆ’10Γ—

We see that only one of the given ordered pairs satisfies the relation π‘₯βˆ’π‘¦=βˆ’10, that is, option D. Hence, the ordered pair (βˆ’12,βˆ’2) satisfies the relation.

In our next example, we see how to find the missing value in an ordered pair, given a linear relation.

Example 2: Completing a Table of Values for a Linear Relation

In the table below some of the π‘₯- and 𝑦-values are missing from the ordered pairs that satisfy the linear relation βˆ’5π‘₯+𝑦=βˆ’3. Find the missing π‘₯ and 𝑦 values.

π‘₯βˆ’403
π‘¦βˆ’1332
Ordered Pair(βˆ’4,β‹―)(β‹―,βˆ’13)(0,β‹―)(3,β‹―)(β‹―,32)

Answer

We can find the missing value in each ordered pair by substituting the known π‘₯- or 𝑦-value into the equation for the linear relation. We then solve for the missing π‘₯- or 𝑦-value.

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 5π‘₯ to both sides, giving 𝑦=5π‘₯βˆ’3.

Now, substituting π‘₯=βˆ’4, we have 𝑦=5Γ—(βˆ’4)βˆ’3. That is, 𝑦=βˆ’20βˆ’3=βˆ’23. Our first ordered pair is therefore (βˆ’4,βˆ’23).

Note that we could have approached this slightly differently by first substituting the known π‘₯-value into the given equation and then solving for 𝑦.

Since for our second ordered pair we know the 𝑦-value, let’s make π‘₯ the subject of our relation βˆ’5π‘₯+𝑦=βˆ’3. We can do this by subtracting 𝑦 from both sides and dividing through by βˆ’5. This gives us π‘₯=35+𝑦5.

Now, substituting 𝑦=βˆ’13 into this equation, we have π‘₯=35+ο€Όβˆ’135.

This evaluates to π‘₯=βˆ’2. Hence, our second ordered pair is (βˆ’2,βˆ’13).

For our third ordered pair, we know that π‘₯=0 and we have 𝑦=5Γ—0βˆ’3=βˆ’3. Hence, the ordered pair is (0,βˆ’3). In our fourth ordered pair, we know that π‘₯=3, hence, 𝑦=5Γ—3βˆ’3=12 and the ordered pair is (3,12). For the last ordered pair, we know that 𝑦=32, hence, π‘₯=35+325=7 and the ordered pair is (7,32).

The missing values from the ordered pairs are therefore 𝑦=βˆ’23, π‘₯=βˆ’2, 𝑦=βˆ’3, 𝑦=12, and π‘₯=7, and we can complete the table as shown.

π‘₯βˆ’4βˆ’2037
π‘¦βˆ’23βˆ’13βˆ’31232
Ordered Pair(βˆ’4,βˆ’23)(βˆ’2,βˆ’13)(0,βˆ’3)(3,12)(7,32)

In our next example, we see how to find an unknown coefficient given a linear relation and an ordered pair satisfying that relation.

Example 3: Finding a Coefficient in a Linear Relation given an Ordered Pair

Given that (βˆ’7,βˆ’3) satisfies the relation 3π‘₯+𝑏𝑦=βˆ’3, find the value of 𝑏.

Answer

To find the value of 𝑏, we substitute the π‘₯- and 𝑦-values from the given ordered pair into the linear relation. We then solve the resulting equation for 𝑏. With π‘₯=βˆ’7 and 𝑦=βˆ’3, the linear relation is 3Γ—(βˆ’7)+𝑏×(βˆ’3)=βˆ’3, which evaluates to βˆ’21βˆ’3𝑏=βˆ’3.

To solve for 𝑏, we add 21 to both sides then divide both sides by βˆ’3 so that 𝑏=βˆ’183=βˆ’6.

The value of 𝑏 that satisfies the given linear relation is therefore 𝑏=βˆ’6.

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 between those points.

For example, the ordered pairs (βˆ’1,βˆ’3) and (2,3) satisfy the linear relation βˆ’2π‘₯+𝑦=βˆ’1 We can represent this relation graphically by plotting the points with coordinates π‘₯=βˆ’1, 𝑦=βˆ’3 and π‘₯=2, 𝑦=3 and drawing a line through those points.

Note that while we have plotted the line corresponding to the linear relation using only two points, in fact, every point on the line is represented by an ordered pair (π‘₯,𝑦) that satisfies the linear relation βˆ’2π‘₯+𝑦=βˆ’1, where π‘₯ and 𝑦 are the coordinates of the point.

With this in mind, looking once again at our postage stamps example, we note that the linear relation 20π‘₯+50𝑦=240, which we can write equivalently as 2π‘₯+5𝑦=24, can be represented by the graph below.

In this case, however, while the linear relation 20π‘₯+50𝑦=240 is represented completely by the plotted line, we know that there are only three points on the line that correspond to the actual scenario of buying stamps with a value of 240 pt. Remember, π‘₯ is the number of 20 pt stamps and 𝑦 is the number of 50 pt stamps, and these cannot be broken down into smaller units. Hence, since the solutions must be positive whole number values, there are only three possibilities: (2,4), (7,2), and (12,0).

In the linear relations discussed so far, that is, relations of the form π‘Žπ‘₯+𝑏𝑦=𝑐, the coefficients π‘Ž and 𝑏 have been nonzero. Let’s now look at the special cases where either π‘Ž or 𝑏 is equal to 0.

  • If π‘Ž=0, our linear relation is of the form 𝑏𝑦=𝑐. Dividing through by the coefficient 𝑏, we have 𝑦=𝑐𝑏. This means that for every value of π‘₯, 𝑦 is equal to the constant 𝑐𝑏. Graphically, this is a horizontal line through the point 𝑦=𝑐𝑏.
  • If 𝑏=0, our linear relation is of the form π‘Žπ‘₯=𝑐. Dividing through by the coefficient π‘Ž, we have π‘₯=π‘π‘Ž. This means that for every value of 𝑦, π‘₯ is equal to the constant π‘π‘Ž. Graphically, this is a vertical line through the point π‘₯=π‘π‘Ž.

In our next two examples, we are asked to plot the graphs of linear relations of this type.

Example 4: Drawing the Graph of a Linear Relation

Given the relation π‘Žπ‘₯+𝑏𝑦=𝑐, sketch its graph if π‘Ž=0, 𝑏=3, and 𝑐=2.

Answer

To sketch the graph of the relation π‘Žπ‘₯+𝑏𝑦=𝑐, where the coefficients take the values given, we first substitute the given values into the equation. This gives 0Γ—π‘₯+3×𝑦=2.

That is, 3𝑦=2. To make 𝑦 the subject of this new equation, we divide through by 3. Hence, 𝑦=23. Another way of looking at this is to recall that if π‘Ž=0 in the linear relation π‘Žπ‘₯+𝑏𝑦=𝑐, then 𝑦=𝑐𝑏. This is exactly what we found by substituting the given coefficients into the relation.

To sketch the graph of the relation 𝑦=23, we note that the interpretation of this relation is that for every value of π‘₯, 𝑦 is equal to the constant value 23. The graph of the relation is therefore a horizontal line through the point 𝑦=23, which we can sketch as shown below.

Example 5: Drawing the Graph of a Linear Relation

Given the relation π‘Žπ‘₯+𝑏𝑦=𝑐, sketch the graph of this relation if π‘Ž=5, 𝑏=0, and 𝑐=1.

Answer

To sketch the graph of the relation π‘Žπ‘₯+𝑏𝑦=𝑐, where the coefficients take the values given, we first substitute the given values into the equation. This gives 5Γ—π‘₯+0×𝑦=1.

That is, 5π‘₯=1, and to make π‘₯ the subject of this new equation, we divide through by 5. We then have π‘₯=15. We can look at this another way by recalling that if 𝑏=0 in the linear relation π‘Žπ‘₯+𝑏𝑦=𝑐, then π‘₯=π‘π‘Ž. In our case, with π‘Ž=5 and 𝑐=1, this is exactly what we found.

To sketch the graph of the relation π‘₯=15, we note that we interpret this relation as follows: for every value of 𝑦, π‘₯ is equal to 15. The graph of the relation is therefore a vertical line through the point π‘₯=15, which we can sketch as shown below, noting that as a decimal, 15=0.2.

In our final example, we are given a linear relation and must find the missing values in a table of ordered pairs. We then use these values to sketch a graph of the relation.

Example 6: Completing a Table of Values for a Linear Relation and Then Graphing the Relation

  • Given the linear relation βˆ’4π‘₯+2𝑦=βˆ’6, complete the table of values below.
    π‘₯015
    π‘¦βˆ’7βˆ’13
  • Using the completed table, sketch a graph to represent the relation.

Answer

Each column in the table represents an ordered pair (π‘₯,𝑦) whose π‘₯- and 𝑦-values together satisfy the given relation βˆ’4π‘₯+2𝑦=βˆ’6. To find the missing values, we substitute the known value of π‘₯ or 𝑦 into the equation for the relation and solve for the unknown value.

For the first ordered pair, we are given that 𝑦=βˆ’7. To find the corresponding π‘₯-value, we substitute 𝑦=βˆ’7 into the given relation as follows: βˆ’4π‘₯+2Γ—(βˆ’7)=βˆ’6βˆ’4π‘₯βˆ’14=βˆ’6.

Now, to make π‘₯ the subject, we add 14 to both sides and divide by βˆ’4, giving π‘₯=βˆ’84=βˆ’2.

Our first missing value is therefore π‘₯=2, which we enter into our table.

π‘₯βˆ’2015
π‘¦βˆ’7βˆ’13

The next missing value is the 𝑦-value corresponding to π‘₯=0. To find this 𝑦-value, we substitute π‘₯=0 into our relation, giving βˆ’4Γ—0+2𝑦=βˆ’62𝑦=βˆ’6.

Now, to make 𝑦 the subject, we divide both sides by 2, and we find 𝑦=βˆ’62=βˆ’3.

Our second missing value is therefore 𝑦=βˆ’3.

The third missing value is the π‘₯-value corresponding to 𝑦=3. Substituting 𝑦=3 into our relation gives βˆ’4π‘₯+2Γ—3=βˆ’6βˆ’4π‘₯+6=βˆ’6.

Solving this for π‘₯, we find π‘₯=3. Our final missing value is the 𝑦-value corresponding to π‘₯=5, and substituting π‘₯=5 into our relation gives βˆ’4Γ—5+2𝑦=βˆ’6βˆ’20+2𝑦=βˆ’6.

Solving this for 𝑦, we find 𝑦=7. We can now complete the table with the missing values we have found.

π‘₯βˆ’20135
π‘¦βˆ’7βˆ’3βˆ’137

We are asked to use the completed table to sketch the graph of the given relation. To do this, we note the ordered pairs of π‘₯- and 𝑦-values, as these are the coordinates of points on the line representing the linear relation. These are (βˆ’2,βˆ’7), (0,βˆ’3), (1,βˆ’1), (3,3), and (5,7). Plotting these points and sketching the line that passes through them gives us the graph of the linear relation βˆ’4π‘₯+2𝑦=βˆ’6, as shown below.

Note that since each term in the given equation has a common factor of 2, the equation can be divided through by 2 without changing the ordered pairs that satisfy the equation. This gives βˆ’2π‘₯+𝑦=βˆ’3, and since there are now no common factors, we say that this is the simplest form of the linear relation. The graph is the same as that of the original relation.

We complete this explainer by recapping the key points covered.

Key Points

  • A linear relation of the form π‘Žπ‘₯+𝑏𝑦=𝑐 can be represented by a set of ordered pairs (π‘₯,𝑦), where each pair of values satisfies the equation of the given linear relation.
  • We may find the value of an unknown coefficient in a linear relation π‘Žπ‘₯+𝑏𝑦=𝑐 given an ordered pair satisfying the linear relation.
  • To sketch the graph of a linear relation, we plot the coordinates corresponding to two or more ordered pairs that satisfy the relation. The line that passes through these points represents the linear relation.
  • If the equation of a linear relation is multiplied by a constant, both the graph of the relation and the ordered pairs satisfying the relation remain unchanged. If there are no common factors in the terms of the equation of the relation, then the linear relation is said to be in its simplest form.
  • If π‘Ž=0 in the linear relation π‘Žπ‘₯+𝑏𝑦=𝑐, then the graph of the relation is the horizontal line passing through the point 𝑦=𝑐𝑏.
  • If 𝑏=0 in the linear relation π‘Žπ‘₯+𝑏𝑦=𝑐, then the graph of the relation is the vertical line passing through the point π‘₯=π‘π‘Ž.

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