Comparing Linear Relations

Which phone plan is cheaper? When will your little brother catch up to you on his bike? Questions like these compare two linear relations. The key is to find the moment when they’re equal: before that point one relation is ahead, and after it the other one is. You can find that point with a table, a graph, or a little algebra.

Key ideas

The big question: when are they equal?

Suppose two relations are written as $y = ax + b$ and $y = cx + d$. The most important point is where they have the same $y$-value for the same $x$-value. On a graph, that’s where the two lines cross: the point of intersection.

In a real situation, the point of intersection tells you when two plans cost the same, when two people are at the same place, or when two amounts of money are equal.

Comparing with tables

Make a table for each relation using the same $x$-values, side by side. Look for the $x$-value where the $y$-values match. Watch which relation is bigger before and after that point.

Tables are quick, but they only show the values you calculate. If the relations are equal at $x = 4.5$ and you only checked whole numbers, you’ll see them “switch places” between $4$ and $5$ without hitting the exact point.

Comparing with graphs

Graph both lines on the same grid. The point where they cross is the point of intersection. On one side of it, one line is lower; on the other side, the other line is lower. A graph shows the whole picture at a glance, but reading the exact point can be tricky, so check it with algebra.

Comparing with algebra: the comparison method

If both relations are solved for $y$, the $y$-values are equal at the intersection, so set the two expressions equal to each other:

$$ax + b = cx + d$$

1. Solve this equation for $x$ (see solving linear equations).
2. Substitute that $x$ into either equation to find $y$.
3. Check the point in the other equation.
4. Say what the point means in the situation.

Parallel lines never meet

If two relations have the same rate of change but different initial values, their lines are parallel. They never cross, so the relations are never equal: the one that starts higher stays higher forever. When you try the comparison method, the $x$-terms cancel and you’re left with something false, like $40 = 25$.

If they have the same rate of change and the same initial value, they’re really the same relation, and the lines lie on top of each other.

In Grade 10 you’ll study this in more depth as solving linear systems by graphing.

Worked examples

Example 1: Two phone plans, with a table and a graph

Plan A costs $25 a month plus $5 per GB of data. Plan B costs $10 a month plus $8 per GB. Which plan is cheaper, and when?

Solution. Let $g$ be the data used (in GB) and $C$ the monthly cost (in dollars).

$$\text{Plan A: } C = 5g + 25 \qquad \text{Plan B: } C = 8g + 10$$

Make a table:

$g$ (GB)	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
Plan A ($)	$25$	$30$	$35$	$40$	$45$	$50$	$55$	$60$
Plan B ($)	$10$	$18$	$26$	$34$	$42$	$50$	$58$	$66$

Both plans cost $50 at $5$ GB. Before that, Plan B is cheaper; after that, Plan A is cheaper. The graph shows the same thing.

The plans cost the same at $(5, 50)$. Plan B starts cheaper, but its higher rate of change makes it more expensive after $5$ GB.

Answer: Plan B is cheaper if you use less than $5$ GB a month. Plan A is cheaper if you use more than $5$ GB. At exactly $5$ GB, both cost $50.

Example 2: Two gyms, with the comparison method

Gym X charges a $60 joining fee plus $25 per month. Gym Y has no joining fee but charges $35 per month. After how many months is the total cost the same? Which gym is cheaper for a year?

Solution. Let $m$ be the number of months and $C$ the total cost in dollars.

$$\text{Gym X: } C = 25m + 60 \qquad \text{Gym Y: } C = 35m$$

Set the expressions equal and solve:

$$\begin{aligned} 25m + 60 &= 35m \\ 60 &= 10m && \text{subtract } 25m \\ 6 &= m && \text{divide by } 10 \end{aligned}$$

Find the cost: $C = 35(6) = 210$. Check in the other equation: $25(6) + 60 = 150 + 60 = 210$. ✓

The two gyms cost the same, $210, after $6$ months. Before that, Gym Y is cheaper (no joining fee). After that, Gym X is cheaper (lower monthly rate).

For a year ($m = 12$): Gym X costs $25(12) + 60 = 360$ dollars and Gym Y costs $35(12) = 420$ dollars. Gym X is cheaper by $60.

Example 3: Catching up

Lin is walking to the library at $80$ m per minute, and she is already $600$ m from home. At that moment, her brother leaves home on his bike, riding the same way at $200$ m per minute. When and where does he catch up?

Solution. Let $t$ be the time in minutes and $d$ the distance from home in metres.

$$\text{Lin: } d = 80t + 600 \qquad \text{Brother: } d = 200t$$

He catches up when they are the same distance from home:

$$\begin{aligned} 200t &= 80t + 600 \\ 120t &= 600 && \text{subtract } 80t \\ t &= 5 && \text{divide by } 120 \end{aligned}$$

Distance: $d = 200(5) = 1000$. Check: $80(5) + 600 = 400 + 600 = 1000$. ✓

He catches up after $5$ minutes, $1000$ m (or $1$ km) from home.

Lin has a $600$ m head start, but her brother gains $120$ m every minute, so he catches up at $(5, 1000)$.

Here’s another way to see it: the brother gains $200 - 80 = 120$ m on Lin every minute. To close a $600$ m gap takes $600 \div 120 = 5$ minutes.

Example 4: Parallel lines

Ana has $40 saved and adds $15 every week. Ben has $25 saved and also adds $15 every week. When will they have the same amount?

Solution. Let $w$ be the number of weeks and $S$ the savings in dollars.

$$\text{Ana: } S = 15w + 40 \qquad \text{Ben: } S = 15w + 25$$

Try the comparison method:

$$\begin{aligned} 15w + 40 &= 15w + 25 \\ 40 &= 25 && \text{subtract } 15w \end{aligned}$$

That’s false, no matter what $w$ is. So there’s no point of intersection: they will never have the same amount. Both save at the same rate, so their lines are parallel, and Ana stays $15 ahead every single week.

Common mistakes

Stopping at the point of intersection. "$g = 5$" isn’t a complete answer to “which plan is cheaper?”. Say what happens on each side of the intersection: Plan B is cheaper below $5$ GB, Plan A above.

Assuming the lower rate is always cheaper. Plan A has the lower rate ($5 per GB vs $8), but it’s more expensive for small amounts of data because of its higher starting fee. You need both the rate and the initial value.

Forgetting to find $y$. After solving for $x$, substitute back to find $y$ (the cost, the distance, the amount). Then check it in the other equation to catch mistakes.

Trusting a rough graph too much. If the lines cross at $(4.6, 37.8)$, a hand-drawn graph might suggest $(5, 38)$. Use the graph to see the big picture, and use algebra for the exact point.

Thinking “no solution” means you made a mistake. If the $x$-terms cancel and you get a false statement like $40 = 25$, the lines are parallel and the relations are never equal. That’s a real answer.

Mixing units. If one plan charges per minute and another per hour, or one price is in cents and the other in dollars, convert first so both relations use the same units.

Practice

1. (Warm-up) Use the comparison method to find the point of intersection of $y = 3x + 4$ and $y = x + 10$.

Solution

$$\begin{aligned} 3x + 4 &= x + 10 \\ 2x &= 6 \\ x &= 3 \end{aligned}$$

Then $y = 3 + 10 = 13$. Check: $3(3) + 4 = 13$. ✓ The point of intersection is $(3, 13)$.

2. (Warm-up) Make a table of values for $y = 2x + 1$ and $y = 4x - 5$ for $x = 0, 1, 2, 3, 4$. Where do the lines meet?

Solution

$x$	$0$	$1$	$2$	$3$	$4$
$y = 2x + 1$	$1$	$3$	$5$	$7$	$9$
$y = 4x - 5$	$-5$	$-1$	$3$	$7$	$11$

The $y$-values match at $x = 3$, so the lines meet at $(3, 7)$.

3. (Warm-up) Will the lines $y = 6x + 2$ and $y = 6x - 1$ ever meet? Explain.

Solution

No. They have the same slope ($6$) but different $y$-intercepts ($2$ and $-1$), so they’re parallel. The first line is always $3$ units above the second.

4. (Core) Bowling Alley A charges $4 for shoes plus $6 per game. Bowling Alley B includes shoes but charges $8 per game.

• (a) For how many games do they cost the same? What is that cost?
• (b) Which is cheaper for $5$ games?

Solution

(a) Let $g$ be the number of games. A: $C = 6g + 4$. B: $C = 8g$.

$$6g + 4 = 8g \quad\Rightarrow\quad 4 = 2g \quad\Rightarrow\quad g = 2$$

Cost: $8(2) = 16$. Check: $6(2) + 4 = 16$. ✓ They both cost $16 for $2$ games.

(b) A: $6(5) + 4 = 34$ dollars. B: $8(5) = 40$ dollars. Alley A is cheaper for $5$ games.

5. (Core) Car rental company P charges $45 per day plus $0.20 per kilometre. Company Q charges $30 per day plus $0.35 per kilometre. For a one-day rental:

• (a) At what distance do they cost the same?
• (b) Which company is cheaper for a $250$ km trip, and by how much?

Solution

(a) Let $k$ be the distance in kilometres. P: $C = 0.20k + 45$. Q: $C = 0.35k + 30$.

$$\begin{aligned} 0.20k + 45 &= 0.35k + 30 \\ 15 &= 0.15k \\ k &= 100 \end{aligned}$$

Cost: $0.20(100) + 45 = 65$. Check: $0.35(100) + 30 = 65$. ✓ They both cost $65 at $100$ km.

(b) P: $0.20(250) + 45 = 50 + 45 = 95$ dollars. Q: $0.35(250) + 30 = 87.5 + 30 = 117.5$ dollars. Company P is cheaper by $117.50 - 95 = 22.50$, so by $22.50.

6. (Core) Candle A is $30$ cm tall and burns down $2$ cm per hour. Candle B is $20$ cm tall and burns down $1$ cm per hour. They’re lit at the same time. When are they the same height, and how tall are they then?

Solution

Let $t$ be the time in hours and $h$ the height in centimetres. A: $h = 30 - 2t$. B: $h = 20 - t$.

$$\begin{aligned} 30 - 2t &= 20 - t \\ 10 &= t && \text{add } 2t \text{, subtract } 20 \end{aligned}$$

Height: $20 - 10 = 10$. Check: $30 - 2(10) = 10$. ✓ After $10$ hours, both candles are $10$ cm tall. (After that, Candle A is shorter, and it burns out first, at $15$ hours. Candle B lasts $20$ hours.)

7. (Core) Sara runs at $4$ m/s. Her younger brother runs at $2.5$ m/s, and she gives him a $60$ m head start in a $150$ m race.

• (a) If the race were long enough, when and where would Sara catch him?
• (b) Who wins the $150$ m race?

Solution

(a) Let $t$ be the time in seconds and $d$ the distance from the start line in metres. Sara: $d = 4t$. Brother: $d = 2.5t + 60$.

$$4t = 2.5t + 60 \quad\Rightarrow\quad 1.5t = 60 \quad\Rightarrow\quad t = 40$$

$d = 4(40) = 160$. Check: $2.5(40) + 60 = 160$. ✓ Sara would catch him after $40$ s, $160$ m from the start.

(b) The catch-up point is at $160$ m, past the $150$ m finish line, so her brother wins. Check the finish times: Sara needs $150 \div 4 = 37.5$ s. Her brother needs $2.5t + 60 = 150$, so $2.5t = 90$ and $t = 36$ s. He finishes first.

8. (Challenge)

• (a) Find the value of $b$ so that $y = 3x + b$ and $y = x + 7$ meet when $x = 2$.
• (b) Find the value of $a$ so that $y = ax + 1$ and $y = 4x - 3$ never meet.

Solution

(a) At $x = 2$, the second line has $y = 2 + 7 = 9$. The first line must also pass through $(2, 9)$:

$$9 = 3(2) + b \quad\Rightarrow\quad b = 3$$

So $b = 3$. Check: $3(2) + 3 = 9$. ✓

(b) The lines never meet if they’re parallel, which means the same slope with different $y$-intercepts. The $y$-intercepts are already different ($1$ and $-3$), so $a = 4$.

9. (Challenge) Kai has $200 and spends $15 per week. Mia has $50 and saves $10 per week. After how many weeks will they have the same amount? How much will each have?

Solution

Let $w$ be the number of weeks and $M$ the money in dollars. Kai: $M = 200 - 15w$. Mia: $M = 50 + 10w$.

$$\begin{aligned} 200 - 15w &= 50 + 10w \\ 150 &= 25w && \text{add } 15w \text{, subtract } 50 \\ w &= 6 \end{aligned}$$

Kai: $200 - 15(6) = 200 - 90 = 110$. Mia: $50 + 10(6) = 50 + 60 = 110$. ✓

After $6$ weeks, they each have $110. The gap of $150 closes by $15 + 10 = 25$ dollars a week, and $150 \div 25 = 6$.