Grade 10 Linear Systems Word Problems: Substitution and Elimination Explained

To solve a linear systems word problem, define two variables, translate two pieces of information into equations, solve the equations together, and check that the answer makes sense in the original situation. The algebra is often shorter than the setup.

This original practice resource focuses on Grade 10 linear systems. For a review of slope, intercepts and equations of lines, start with our linear relations guide.

A reliable four-step method

• 1. Define each variable with its meaning and units.
• 2. Write one equation for each independent condition. A total number and a total cost usually produce different equations.
• 3. Choose substitution or elimination and show your steps.
• 4. Substitute the result into both original equations, then write a sentence answering the question.

Use substitution when a variable is already isolated or has coefficient 1. Use elimination when adding or subtracting the equations will cancel a variable, perhaps after multiplication. Both methods should give the same solution for a system with one intersection.

Eight practice problems

Try each setup before reading its solution. Prices are fictional and exclude any extra fees or taxes unless stated.

1. Tickets to a school performance

A school sells 80 tickets. Adult tickets cost $12 and student tickets cost $7. Total ticket revenue is $710. How many of each were sold?

Solution: Let a be adult tickets and s be student tickets. The equations are a + s = 80 and 12a + 7s = 710. Multiply the first equation by 7 to get 7a + 7s = 560. Subtract it from the revenue equation: 5a = 150, so a = 30. Then s = 80 − 30 = 50.

Check: 30 + 50 = 80 and 12(30) + 7(50) = 360 + 350 = 710. The school sold 30 adult tickets and 50 student tickets.

2. Two mobile plans

Plan A costs $20 per month plus $3 per gigabyte used. Plan B costs $35 per month plus $1.50 per gigabyte. At what usage do they cost the same, and what is that cost?

Solution: Let g be gigabytes and C be monthly cost in dollars. Write C = 20 + 3g and C = 35 + 1.5g. Substitute by setting the right sides equal: 20 + 3g = 35 + 1.5g. Then 1.5g = 15, so g = 10. Substituting gives C = 20 + 30 = $50.

Below 10 GB, Plan A is cheaper under this model; above 10 GB, Plan B is cheaper. The conclusion assumes these pricing rules continue to apply.

3. A rectangular garden

A garden has perimeter 54 metres. Its length is 3 metres more than twice its width. Find its dimensions.

Solution: Let L and W be the length and width in metres. Write 2L + 2W = 54 and L = 2W + 3. Divide the perimeter equation by 2: L + W = 27. Substitute: 2W + 3 + W = 27, so 3W = 24 and W = 8. Then L = 19.

Check: 2(19) + 2(8) = 54. The garden measures 19 m by 8 m. A frequent error is using L + W = 54 and forgetting that a rectangle has two lengths and two widths.

4. Notebooks and pens

Three notebooks and two pens cost $16. Two notebooks and five pens cost $18. Find the price of each item.

Solution: Let n and p be the prices in dollars. Write 3n + 2p = 16 and 2n + 5p = 18. Multiply the first equation by 5 and the second by 2: 15n + 10p = 80 and 4n + 10p = 36. Subtract: 11n = 44, so n = 4. Substitute into the first equation: 12 + 2p = 16, giving p = 2.

Check the second purchase: 2(4) + 5(2) = 18. A notebook costs $4 and a pen costs $2.

5. Catching up on a path

Walker A starts 300 metres ahead of Walker B. Both move in the same direction at constant speeds. Walker A travels at 60 metres per minute, and Walker B at 90 metres per minute. When does B catch A?

Solution: Measure distance d from B’s starting point, and let t be minutes after they begin. Write d = 300 + 60t for A and d = 90t for B. At the meeting point, 300 + 60t = 90t, so 300 = 30t and t = 10 minutes. The meeting point is d = 90(10) = 900 metres from B’s starting point.

Check A’s position: 300 + 60(10) = 900. Using a shared origin is essential; otherwise the two distance expressions cannot be compared directly.

6. Coins in a jar

A jar contains 32 coins, all quarters and loonies. Their total value is $20. How many of each coin are there?

Solution: Let q be quarters and l be loonies. Write q + l = 32 and 0.25q + l = 20. Subtract the value equation from the number equation to eliminate l: 0.75q = 12. Thus q = 16 and l = 16.

Check the value: 16($0.25) + 16($1) = $4 + $16 = $20. There are 16 quarters and 16 loonies. Keep dollar and cent values consistent when forming the equations.

7. Two parallel cost models

A printing service charges C = 10 + 0.08p dollars for p pages. Another charges C = 15 + 0.08p. When do the costs match?

Solution: Setting the costs equal gives 10 + 0.08p = 15 + 0.08p. Subtracting 0.08p leaves 10 = 15, which is false. There is no solution. The lines have the same slope and different intercepts. The second service costs $5 more at every page count in the model.

No solution is a valid conclusion; do not force a page count from a contradictory equation.

8. When two statements are the same condition

A snack order contains sandwiches and drinks. One statement says 2s + d = 18. A second says 4s + 2d = 36. Can you determine a unique pair (s, d)?

Solution: The second equation is twice the first, so it provides no new independent information. As a system over the real numbers, it has infinitely many solutions on the line d = 18 − 2s.

If s and d are non-negative whole-number item counts, the context restricts the possibilities to 10 pairs, from (0, 18) to (9, 0). There is still no unique order. A genuinely independent condition is needed. This distinction between algebraic solutions and meaningful contextual answers is part of checking a model.

Common mistakes and how to catch them

• Adding unlike quantities: The number of tickets and their dollar value belong in separate equations.
• Losing a negative sign: When subtracting equations, subtract every term on the second line.
• Forgetting units: Define time in minutes or hours before using a speed.
• Accepting impossible counts: A negative number of tickets or a fractional number of coins signals a setup error or incompatible information.
• Checking only one equation: A system’s solution must satisfy both.

Build confidence with the setup

For related practice, read our Grade 10 quadratic equations guide. If translating words into equations is the difficult step, explore Grade 10 math tutoring or book a free one-hour assessment.

Curriculum reference: Ontario Grades 9–10 Mathematics, Grade 10 analytic geometry and linear systems. These are original practice problems, not an official assessment.