Representing Linear Relations

A linear relation can be shown as a real situation, a table of values, a graph, or an equation. They all tell the same story, just in different languages. This page shows you how to move between them, and how to find the two numbers that describe every linear relation: the rate of change and the initial value.

Key ideas

One relation, four representations

Here’s a taxi fare: $4 to start, plus $2 for every kilometre.

Table of values. Let $d$ be the distance in kilometres and $C$ the cost in dollars.

$d$ (km)	$0$	$1$	$2$	$3$	$4$	$5$
$C$ (dollars)	$4$	$6$	$8$	$10$	$12$	$14$

Equation. Start with $4$, then add $2$ for each kilometre:

$$C = 2d + 4$$

Graph. Plot the points from the table and join them with a straight line (the blue line in the figure below). It’s a line because the cost goes up by the same amount every kilometre.

Rate of change

The rate of change is how much $y$ changes when $x$ goes up by $1$. For the taxi it’s $2 per kilometre. You can find it in every representation:

Representation	Where to find the rate of change	Taxi
situation	the amount “per” or “for each”	$2 per km
table	the first difference (when $x$ goes up by $1$)	$6 - 4 = 2$
graph	the slope, $\dfrac{\text{rise}}{\text{run}}$	$\dfrac{4}{2} = 2$
equation	the number multiplying $x$	the $2$ in $C = 2d + 4$

A rate of change has units: here it’s dollars per kilometre. If it’s negative, the relation goes down as $x$ goes up. (For more on slope, including a formula for any two points, see equations of lines.)

Initial value

The initial value is the value of $y$ when $x = 0$. It’s where the relation starts.

Representation	Where to find the initial value	Taxi
situation	the starting amount or fixed fee	$4 to start
table	the $y$-value in the $x = 0$ column	$4$
graph	the $y$-intercept (where the line crosses the $y$-axis)	$(0, 4)$
equation	the number added on	the $4$ in $C = 2d + 4$

The equation y = mx + b

Every linear relation can be written as

$$y = mx + b$$

where $m$ is the rate of change (the slope) and $b$ is the initial value (the $y$-intercept). The Grade 9 curriculum also writes this as $y = ax + b$; the letters are different but the idea is the same.

In real situations we often use letters that remind us what they stand for, like $C$ for cost and $d$ for distance.

Direct and partial variation

• Direct variation: the initial value is $0$, so the equation is $y = mx$. The graph goes through the origin $(0, 0)$. If $x$ doubles, $y$ doubles too.
• Partial variation: the initial value is not $0$, so the equation is $y = mx + b$ with $b \ne 0$. The graph crosses the $y$-axis somewhere other than the origin. There’s a fixed part ($b$) and a part that varies ($mx$).

Suppose a second taxi company charges $2 per kilometre with no starting fee. Its cost is $C = 2d$, a direct variation. Our first taxi, $C = 2d + 4$, is a partial variation.

$C = 2d + 4$ (partial variation) and $C = 2d$ (direct variation) have the same rate of change, so the lines are parallel. Only the starting point is different.

Worked examples

Example 1: From a situation to an equation

Use the taxi fare $C = 2d + 4$ from above.

• (a) How much does a $7.5$ km ride cost?
• (b) How far can you go for $30?

Solution.

(a) Substitute $d = 7.5$:

$$C = 2(7.5) + 4 = 15 + 4 = 19$$

The ride costs $19.

(b) Substitute $C = 30$ and solve for $d$:

$$\begin{aligned} 30 &= 2d + 4 \\ 26 &= 2d && \text{subtract } 4 \\ 13 &= d && \text{divide by } 2 \end{aligned}$$

You can go $13$ km. Check: $2(13) + 4 = 26 + 4 = 30$. ✓

Example 2: From a table to an equation

A phone plan’s monthly cost depends on how much data you use.

Data, $g$ (GB)	$0$	$1$	$2$	$3$	$4$
Cost, $C$ (dollars)	$30$	$35$	$40$	$45$	$50$

• (a) Find the rate of change and the initial value, and say what each means.
• (b) Write the equation.
• (c) How much does the plan cost in a month when you use $9$ GB?

Solution.

(a) The first differences are all $5$, so the relation is linear with a rate of change of $5 per GB: each gigabyte adds $5. When $g = 0$, $C = 30$, so the initial value is $30: the base price of the plan, even if you use no data.

(b) $C = 5g + 30$. This is a partial variation.

(c) $C = 5(9) + 30 = 45 + 30 = 75$. The plan costs $75.

Example 3: From a graph to an equation

The graph shows Sam’s distance from home while walking to a friend’s house.

Sam’s distance from home, $d$ metres, after $t$ minutes.

• (a) Find the rate of change and the initial value. What do they mean?
• (b) Write an equation for the relation.
• (c) Sam’s friend lives $1000$ m from Sam’s home. When does Sam arrive?

Solution.

(a) Use the two marked points, $(0, 200)$ and $(5, 600)$. From $t = 0$ to $t = 5$, the distance goes from $200$ m to $600$ m:

$$\text{rate of change} = \frac{\text{rise}}{\text{run}} = \frac{600 - 200}{5 - 0} = \frac{400}{5} = 80$$

Sam walks at $80$ metres per minute. The line crosses the $d$-axis at $200$, so the initial value is $200$ m: Sam was already $200$ m from home when the timing started.

(b) $d = 80t + 200$

(c) Solve $80t + 200 = 1000$: $80t = 800$, so $t = 10$. Sam arrives after $10$ minutes.

Check: $80(10) + 200 = 800 + 200 = 1000$. ✓

Example 4: Direct or partial?

For each situation, write an equation and say whether it’s a direct or partial variation.

• (a) Gas costs $1.50 per litre. Let $C$ be the cost of $L$ litres.
• (b) Bowling costs $5 to rent shoes plus $7 per game. Let $C$ be the cost of $g$ games.

Solution.

(a) There’s no fixed fee, so $C = 1.5L$. This is a direct variation. Check the doubling rule: $10$ L costs $15 and $20$ L costs $30, twice as much. ✓

(b) The shoe rental is a fixed part and the games are the varying part, so $C = 7g + 5$. This is a partial variation. Doubling doesn’t work here: $2$ games cost $19 but $4$ games cost $33, not $38.

Common mistakes

Using the first $y$-value as the initial value when the table doesn’t start at $x = 0$. If a table starts at $x = 2$, the first $y$-value is not $b$. Work backwards using the rate of change, or substitute a point into $y = mx + b$ and solve for $b$ (see Practice question 4).

Forgetting to divide when $x$ goes up by more than $1$. If $x$ goes up by $2$ and $y$ goes up by $10$, the rate of change is $10 \div 2 = 5$, not $10$.

Mixing up $m$ and $b$. For “$4 to start plus $2 per km”, the equation is $C = 2d + 4$, not $C = 4d + 2$. The “per” amount multiplies the variable; the fixed amount is added on.

Thinking every line passes through the origin. Only direct variations do. If there’s a starting fee or a head start, the graph crosses the $y$-axis above (or below) the origin.

Leaving out units. “The rate of change is $80$” doesn’t say much. ”$80$ metres per minute” does. Units also help you check that your answer makes sense.

Misreading the scale on a graph. In Example 3, each grid line on the vertical axis is $100$ m, not $1$ m. Always read the axis labels before counting squares.

Practice

1. (Warm-up) For $y = 6x + 15$, state the rate of change and the initial value. Is it a direct or partial variation?

Solution

The rate of change is $6$ and the initial value is $15$. Since the initial value isn’t $0$, it’s a partial variation.

2. (Warm-up) Write an equation for this table. Is it a direct or partial variation?

$x$	$0$	$1$	$2$	$3$
$y$	$0$	$7$	$14$	$21$

Solution

The first differences are all $7$, and $y = 0$ when $x = 0$. So $y = 7x$, a direct variation.

3. (Warm-up) Renting a canoe costs $20 plus $12 per hour. Write an equation for the cost $C$ of renting for $h$ hours, and find the cost of a $3$-hour rental.

Solution

$C = 12h + 20$.

For $h = 3$: $C = 12(3) + 20 = 36 + 20 = 56$. A $3$-hour rental costs $56.

4. (Core) Find the equation of the linear relation in this table.

$x$	$2$	$3$	$4$	$5$
$y$	$11$	$14$	$17$	$20$

Solution

The first differences are all $3$, so $m = 3$. The table doesn’t start at $x = 0$, so find $b$ by substituting the point $(2, 11)$ into $y = 3x + b$:

$$11 = 3(2) + b \quad\Rightarrow\quad 11 = 6 + b \quad\Rightarrow\quad b = 5$$

The equation is $y = 3x + 5$. (You can also work backwards: from $x = 2$ to $x = 0$ is two steps back, so $11 - 3 - 3 = 5$.)

Check: $x = 5$ gives $3(5) + 5 = 20$. ✓

5. (Core) A water tank holds $50$ L. When a tap is opened, the line on a graph of volume against time goes through $(0, 50)$ and $(4, 30)$, where time is in minutes.

• (a) Find the rate of change, with units. What does its sign tell you?
• (b) Write an equation for the volume $V$ after $t$ minutes.
• (c) When will the tank be empty?

Solution

(a)

$$\text{rate of change} = \frac{30 - 50}{4 - 0} = \frac{-20}{4} = -5$$

The rate of change is $-5$ litres per minute. It’s negative because the volume goes down over time.

(b) The initial value is $50$, so $V = -5t + 50$ (or $V = 50 - 5t$).

(c) Solve $50 - 5t = 0$: $5t = 50$, so $t = 10$. The tank is empty after $10$ minutes.

6. (Core) A pattern is made of squares. Figure 1 has $6$ squares, Figure 2 has $10$, and Figure 3 has $14$.

• (a) Write an equation for the number of squares $s$ in Figure $n$.
• (b) How many squares are in Figure 20?
• (c) Is this a direct or partial variation?

Solution

(a) The pattern adds $4$ squares each time, so $m = 4$. Going back one step from Figure 1 gives a “Figure 0” with $6 - 4 = 2$ squares, so $b = 2$:

$$s = 4n + 2$$

Check: $n = 3$ gives $4(3) + 2 = 14$. ✓

(b) $s = 4(20) + 2 = 82$ squares.

(c) Partial variation, because the initial value is $2$, not $0$.

7. (Core) A part-time job pays $17.50 per hour.

• (a) Write an equation for your pay $P$ for working $h$ hours.
• (b) How much do you earn in a week where you work $12$ hours?
• (c) Explain why this is a direct variation.

Solution

(a) $P = 17.5h$

(b) $P = 17.5(12) = 210$. You earn $210.

(c) There’s no fixed amount: working $0$ hours earns $0, so the graph goes through the origin. Doubling the hours doubles the pay: $24$ hours would earn $17.5(24) = 420$ dollars, which is $2 \times 210$.

8. (Challenge) A long-distance calling plan has this cost table.

Minutes, $x$	$0$	$10$	$20$	$30$
Cost, $C$ (dollars)	$12.00$	$14.50$	$17.00$	$19.50$

• (a) Find the rate of change per minute, and write the equation.
• (b) How much does $90$ minutes of calling cost?
• (c) How many minutes can you talk for $40?

Solution

(a) Each $10$ minutes adds $14.50 - 12.00 = 2.50$ dollars. Per minute, that’s $2.50 \div 10 = 0.25$, so $0.25 per minute. The initial value is $12, so

$$C = 0.25x + 12$$

(b) $C = 0.25(90) + 12 = 22.5 + 12 = 34.5$. It costs $34.50.

(c) Solve $0.25x + 12 = 40$: $0.25x = 28$, so $x = 28 \div 0.25 = 112$. You can talk for $112$ minutes.

Check: $0.25(112) + 12 = 28 + 12 = 40$. ✓

9. (Challenge)

• (a) $y$ varies directly with $x$, and $y = 18$ when $x = 4$. Find $y$ when $x = 10$.
• (b) A different linear relation has an initial value of $3$ and also passes through $(4, 18)$. Find $y$ when $x = 10$.

Solution

(a) Direct variation means $y = mx$. Substitute $(4, 18)$: $18 = 4m$, so $m = 4.5$. Then $y = 4.5(10) = 45$.

(b) Now $y = mx + 3$. Substitute $(4, 18)$:

$$18 = 4m + 3 \quad\Rightarrow\quad 15 = 4m \quad\Rightarrow\quad m = 3.75$$

So $y = 3.75x + 3$, and when $x = 10$, $y = 3.75(10) + 3 = 37.5 + 3 = 40.5$.

Both relations pass through $(4, 18)$, but they give different answers at $x = 10$ because they start in different places.