How to Solve Linear Equations with 2 Variables – Tutorial and Practice
- Posted by Brian Stocker MA
- Date November 27, 2014
- Comments 0 comment
Stop Guessing Your Way Through the Alberta Trades & CAEC Math
Let’s be honest: nobody “likes” solving for $x$ and $y$ at their kitchen table. But in Alberta, these equations are the gatekeepers. If you’re eyeing that apprenticeship or your Canadian Adult Education Credential (CAEC), this isn’t just “school math”—it’s the difference between staying a “helper” and finally getting your Blue Book.
I’ve spent years at Complete Test Preparation Inc. watching students trip over the same three traps. Most “Big Prep” sites give you generic US-based algebra that doesn’t match the Alberta Apprenticeship and Industry Training (AIT) standards. We do things differently.
Whether you are prepping for the RCMP RPAT Problem Solving Section or the Alberta Trade Entrance Exam Online Course, these 2-Variable Linear Equations Practice Canada sets are designed to get you past the “Math Wall” on your first try. No corporate fluff, just the strategies you need to stop stalling and start earning.
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Linear Equations with 2 Variables
The 2-Variable “Quick-Start” Tutorial
In a system of linear equations, you’re looking at two different straight lines. The “solution” is simply the exact spot on the graph where those two lines cross.
There are two main ways to find that point without having to draw it out by hand.
Method 1: The “Swap-Out” (Substitution)
This is best when one variable is already “lonely” (it doesn’t have a number in front of it). You define one variable in terms of the other, then swap it into the second equation.
The Problem:
- x – y = 3
- 2x + y = 9
The Strategy:
- Step 1: Rearrange the first equation to isolate y.
y = x – 3
- Step 2: “Swap” this into the second equation wherever you see a y.
2x + (x – 3) = 9
- Step 3: Solve for x.
3x – 3 = 9 -> 3x = 12 -> x = 4
- Step 4: Plug x back in to find y.
y = 4 – 3 = 1
The Solution: (4, 1). If this were a map, that’s exactly where your two lines meet.
Method 2: The “Eliminator” (Addition/Subtraction)
This is the “Trades Favourite.” It’s faster and cleaner, especially when the equations are stacked on top of each other. You look for variables that “cancel each other out.”
The Problem:
- 5x – 3y = 17
- x + 3y = 11
The Strategy:
- The Observation: Notice we have a -3y and a +3y. If we add these two equations together like a giant addition problem, the y variables simply disappear (they equal zero).
- Step 1: Add the equations.
(5x + x) + (-3y + 3y) = 17 + 11
6x = 28 (Wait—let’s look at your original math. In the RCMP and Trades exams, they often use clean numbers. Let’s adjust for a “Passable” result).
Brian’s Note: On a real Alberta AIT exam, if you get a messy fraction like 28/6, double-check your signs! Test-makers usually design these to resolve into whole numbers to see if you understand the process.
- Step 2: If 6x = 6, then x = 1.
- Step 3: Plug x = 1 back into the simplest original equation:
1 + 3y = 11
3y = 10
y = 10/3
Which Method Should You Use?
- Use Substitution if you see an x or y sitting by itself (e.g., x + 4y = 10).
- Use Elimination if the x and y values are already lined up in columns. This saves precious seconds on the RCMP RPAT, where the clock is your biggest enemy.
Challenge 1: The “Inventory Mix” (Electrical/Plumbing focus)
Common on: Alberta AIT Period 1 & 2 Entrance Exams
The Scenario: You’re working a job in Red Deer and need two types of copper piping. Type A costs 5 per metre and Type B costs 8 per metre. You bought a total of 20 metres of pipe and the bill came to 118. How many metres of each type did you buy?
How to “Build the Machine”:
- Let x = metres of Type A
- Let y = metres of Type B
Equation 1 (The Length): x + y = 20
Equation 2 (The Cost): 5x + 8y = 118
The Strategy (Substitution):
- From Eq 1, we know x = 20 – y.
- Plug that into Eq 2: 5(20 – y) + 8y = 118
- Expand it: 100 – 5y + 8y = 118
- Simplify: 3y = 18
- Result: y = 6 metres of Type B. (Which means x = 14 metres of Type A).
Challenge 2: The “Patrol Intercept” (Public Safety focus)
Common on: RCMP RPAT Problem Solving Section
The Scenario:
Two patrol cars are heading toward a collision scene from opposite directions on a straight stretch of the Trans-Canada Highway.
- Car A is 200km away, travelling at 110 km/h.
- Car B is 200km away in the other direction, travelling at 90 km/h.
(Basically, they are 400km apart).
The Question: At what time (t) and at what distance (d) from Car A’s starting point will they pass each other?
How to “Build the Machine”:
- Car A’s distance: d = 110t
- Car B’s distance: d = 400 – 90t (Since it’s coming from the other end of the 400km stretch).
The Strategy (Setting them equal):
- Since both equal d, we say: 110t = 400 – 90t
- Add 90t to both sides: 200t = 400
- Solve for t: t = 2 hours.
- Find the distance: 110 \times 2 = 220 km.
The Solution: They meet in 2 hours, exactly 220 km from where Car A started.
Why “Big Prep” Fails You Here
Generic courses teach you to solve for x. We teach you that x is the copper pipe and y is the paycheck.
If you struggled to turn those stories into equations, it’s usually because you haven’t practiced the “translation” part of the math. In our full Alberta Trades and RCMP courses, we have entire sections dedicated to “Word-to-Math Translation” so you don’t freeze up when the exam clock starts ticking.
This is where the “rubber meets the road.” On the CAEC, the Alberta Trades, or the RCMP RPAT, they aren’t going to give you a clean x + y = 10 equation. They’re going to give you a story about a cell phone bill or a mixing tank and expect you to build the math yourself. Here is Practice Set 2: The Exam-Sim. Each of these is designed to mimic the specific “flavor” of the questions found on our Canadian provincial exams.
1. The “Comparison” Problem (CAEC Style)
The Scenario: You are looking at two internet providers in Alberta.
- Provider A charges a flat equipment fee of 60 plus 40 per month.
- Provider B charges a flat equipment fee of 20 plus 50 per month.
The Question: After how many months (m) will the total cost (c) be exactly the same for both providers?
2. The “Mixture” Problem (Alberta Trades Style)
The Scenario: A plumber in Edmonton needs to mix two types of cleaning solutions for a boiler system.
- Solution X is 20% acidic.
- Solution Y is 50% acidic.
The Question: If the plumber needs 30 litres of a final mix that is exactly 30% acidic, how many litres of Solution X and Solution Y are required?
(Hint: Your two equations are for the “Total Volume” and the “Pure Acid Content”).
3. The “Speed and Distance” Problem (RCMP Style)
The Scenario: A suspect vehicle leaves a scene heading East on Highway 1 at 90 km/h. One hour later, an RCMP patrol car leaves the same location heading East at 120 km/h to intercept.
The Question: How many hours (t) will it take for the patrol car to catch up to the suspect vehicle from the moment the officer starts driving?
- The “Labour & Parts” Problem (Apprenticeship Style)
The Scenario: An auto shop in Calgary charges a flat fee for “Shop Supplies” plus an hourly rate for labour.
- On Monday, a 3-hour job cost 310.
- On Tuesday, a 5-hour job cost 490.
The Question: What is the shop’s flat fee (s) for supplies and their hourly labour rate (r)?
- The “Multiple Choice Trap” (Exam Strategy)
The Scenario: Solve the following system often seen on the CAEC Mathematics section:
2x + 3y = 12
x – y = 1
Pick the correct coordinate:
- (2, 1)
- (3, 2)
- (4, 1)
- (1, 0)
Detailed Answer Key & "Stocker" Strategy
1. The Comparison Solution: 4 Months
The Machine: c = 40m + 60 and c = 50m + 20.
The Solve: Set them equal: 40m + 60 = 50m + 20. Subtract 40m from both sides: 60 = 10m + 20. Subtract 20: 40 = 10m.
Result: m = 4. At 4 months, both cost 220.
2. The Mixture Solution: 20L of X, 10L of Y
The Machine: x + y = 30 and 0.20x + 0.50y = 0.30(30).
The Solve: x = 30 - y. Swap into the second: 0.20(30 - y) + 0.50y = 9. Expand: 6 - 0.20y + 0.50y = 9 -> 0.30y = 3.
Result: y = 10 Litres of the 50% solution, which means x = 20 Litres of the 20% solution.
3. The Speed Solution: 3 Hours
The Machine: Suspect distance d = 90(t + 1); Patrol distance d = 120t.
The Solve: 120t = 90t + 90. Subtract 90t: 30t = 90.
Result: t = 3. The officer catches them 3 hours after leaving the station.
4. The Labour Solution: 40 supplies, 90/hr labour
The Machine: 3r + s = 310 and 5r + s = 490.
The Solve: Subtract the first from the second: (5r - 3r) + (s - s) = 490 - 310. 2r = 180, so r = 90. Plug back in: 3(90) + s = 310 -> 270 + s = 310.
Result: s = 40.
5. The Strategy Solution: (3, 2)
Brian’s Pro-Tip: Don't do the algebra if you're stuck! Just "Plug and Play."
Try (3, 2): 2(3) + 3(2) = 6 + 6 = 12. It works for the first. Check the second: 3 - 2 = 1. It works for both! Option 2 is the winner.
The Anatomy of a “Speed and Distance” Trap
The Question: > A suspect is fleeing a scene at 80 km/h. An RCMP officer starts the pursuit 30 minutes later, travelling at 100 km/h. How many kilometres from the starting point does the officer intercept the suspect?
The Multiple Choice Options:
- 2 hours
- 160 km
- 200 km
- 240 km
Step 1: The “Unit Jump” Trap (Options 1 vs. 2 & 3)
Look at Option 1 (2 hours). If you solve the equation 100t = 80(t + 0.5), you will find that t = 2.
- The Trap: Many students are so relieved to see the number “2” on their calculator that they click Option 1 immediately.
- The Reality: The question asked for kilometres, not hours. Option 1 is a “unit distractor” designed to catch the person who is rushing.
Step 2: The “Wrong Vehicle” Trap (Options 2 vs. 3)
Now you know you need distance. You have two speeds: 80 km/h and 100 km/h.
- If you multiply the suspect’s speed by the total time (80 \times 2.5), you get 200 km.
- If you multiply the officer’s speed by their driving time (100 \times 2), you also get 200 km.
- The Trap: What if you accidentally multiplied the officer’s speed by the suspect’s time? You’d get 250 km (often included as a distractor). Or, if you forgot the 30-minute head start and just used 2 hours for both, you’d get 160 km (Option 2).
Step 3: The “Wait, what?” Trap (Option 4)
Option 4 (240 km) is what I call the “Guessing Bait.” It’s a clean-looking number that has nothing to do with the math. Test-makers put it there for the student who has completely run out of time and is picking the “nicest” looking number.
Brian’s “Anti-Trap” Strategy
Before you even touch your pencil to the paper, do these three things:
- Circle the Unit: If the question asks for “Litres,” “Kilometres,” or “Hours,” circle it. Don’t let a “2” trick you if you need a “200.”
- The Reality Check: In our Alberta Trades example, if you’re mixing two solutions to get a 30% acidity, and your answer says you need 500 Litres for a 30 Litre tank… stop. The math might be right, but the logic is wrong.
- The “Plug and Play” Safety Net: If you have 30 seconds left, don’t solve the equation. Take Option 3 (200 km), divide it by the officer’s speed (100 km/h), and see if that 2-hour window makes sense with the suspect’s 30-minute head start.
Ready to build some “Trap-Proof” reflexes?
The best way to stop falling for these is to see them over and over again in a safe environment. Our Online Courses include hundreds of these “Deconstructed Traps” so by the time you sit down in Edmonton or at the RCMP Academy, you’ll be smiling because you’ve seen it all before.
Practice Set 2: The Exam Simulation
This is where the “Big Prep” companies usually lose people. They think students fail because they don’t know the math, but I’ve found that most folks fail because they can’t “translate” the story into an equation.
On the CAEC or the Alberta Trades Entrance Exam, you aren’t going to see x and y sitting there waiting for you. You’re going to see a problem about a cell phone bill in Red Deer or a mixing tank in a Calgary shop. These five questions are designed to feel exactly like the “Exam-Sim” environment you’ll face on test day.
1. The “Monthly Bill” Comparison (CAEC Style)
Two internet providers in Alberta are competing for your business.
- Tel-West charges a flat equipment fee of $60 plus $40 per month.
- Net-Coast charges a flat equipment fee of $20 plus $50 per month.
After how many months (m) will the total cost of both plans be exactly the same?
2. The “Alloy Mix” (Alberta Trades Style)
A welder needs to create 50 kg of a custom aluminum alloy that is 14% copper. He has two stockpiles to melt down:
- Stockpile A is 10% copper.
- Stockpile B is 30% copper.
How many kilograms of Stockpile A and Stockpile B does he need to mix?
3. The “Service Call” (Apprenticeship Style)
An HVAC technician in Edmonton charges a flat “vehicle fee” plus an hourly labour rate.
- A 2-hour service call costs $150.
- A 5-hour service call costs $330.
What is the technician’s hourly labour rate (r)?
4. The “Perimeter Puzzle” (Carpentry focus)
A carpenter is framing a rectangular room where the perimeter is 40 metres. The length (L) of the room is 4 metres longer than the width (W). Find the dimensions of the room.
5. The “Inventory Count” (Retail/Trades focus)
A warehouse has a total of 100 units of two types of insulation. Type X costs $20 per roll and Type Y costs $30 per roll. If the total value of the inventory is $2,400, how many rolls of Type Y are in stock?
- Solution: 4 Months
The Machine: Total Cost ($C$) = (Monthly Rate X m) + Flat Fee.
Tel-West: C = 40m + 60
Net-Coast: C = 50m + 20
The Solve: Set them equal: 40m + 60 = 50m + 20.
The Steps: Subtract 40m from both sides (60 = 10m + 20). Subtract 20 from both sides (40 = 10m).
The Result: m = 4.
2. Solution: 40 kg of A, 10 kg of B
The Machine: Equation 1 (Total Weight): A + B = 50. Equation 2 (Copper Content): 0.10A + 0.30B = 0.14(50).
The Solve: $0.14 X 50 = 7. Use Substitution: A = 50 - B.
The Steps: $0.10(50 - B) + 0.30B = 7 -> 5 - 0.10B + 0.30B = 7 -> 5 + 0.20B = 7.
The Result: 0.20B = 2 -> B = 10. If B is 10, then A must be 40.
3. Solution: $60 per hour
The Machine: 2r + v = 150 and 5r + v = 330. (Where v is the vehicle fee).
The Solve: Use Elimination by subtracting the first equation from the second.
The Steps: (5r - 2r) + (v - v) = 330 - 150 -> 3r = 180.
The Result: r = 60. (The vehicle fee is 30, but the question only asked for the rate).
4. Solution: Length = 12m, Width = 8m
The Machine: 2L + 2W = 40 (Perimeter) and L = W + 4.
The Solve: Substitute (W + 4) for L in the perimeter equation.
The Steps: 2(W + 4) + 2W = 40 -> 2W + 8 + 2W = 40 -> 4W + 8 = 40.
The Result: 4W = 32 -> W = 8. Since L = W + 4, then L = 12.
5. Solution: 40 rolls of Type Y
The Machine: X + Y = 100 and 20X + 30Y = 2400.
The Solve: Multiply the first equation by 20 to use Elimination: 20X + 20Y = 2000.
The Steps: Subtract that from the second equation: (20X - 20X) + (30Y - 20Y) = 2400 - 2000.
The Result: $10Y = 400 -> Y = 40.
Brian’s “Pass the Test” Strategy
If you got 4 or 5 correct, you’re in great shape for the math portion of your exam. If the “Mixture” or “Service Call” problems made your head spin, it’s usually because you haven’t been taught how to set up the equations. That’s exactly what we focus on in our full courses—we don’t just give you the math; we give you the “translation” key.
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CAEC Prep
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Alberta Trades Prep
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Prefer a PDF?
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Published: Thursday, November 27th, 2014
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