Ontario · MPM2D · Grade 10
Ontario Grade 10 Linear Systems Practice Questions
Two equations, two unknowns — find where they meet. Practise substitution, elimination and graphing. Word problems add one step: name the variables, write the two equations, then solve. Try on paper first, then check the steps.
What you need to know
- Substitution — isolate one variable and substitute.
- Elimination — scale to opposite coefficients, add, then back-substitute.
- Graphing — put each line in y = mx + b and read the intersection.
- Word problems — define variables, write the two equations, solve, state units.
Practice Questions (13)
Question 1 · Linear Systems Concepts
Is the following statement true or false?
"If two equations in a linear system represent the same line, the system has infinitely many solutions."
Show Solution
If the equations represent the same line (coincident lines), they intersect at every point along the line, yielding infinitely many solutions. The answer is a.
Answer: True
Question 2 · Solving by Graphing
Solve the linear system by graphing.
$$y = 2x - 1$$
$$y = -x + 5$$
What is the point of intersection?
Show Solution
Graph y = 2x - 1: y-intercept at (0,-1), slope 2.
Graph y = -x + 5: y-intercept at (0,5), slope -1.
The lines intersect at (2, 3).
Check: 3 = 2(2) - 1 = 3 ✓
3 = -(2) + 5 = 3 ✓
Answer: (2,3)
Question 3 · Solving by Graphing
Solve the linear system by graphing.
$$y = \frac{1}{2}x + 2$$
$$y = 2x - 1$$
What is the point of intersection?
Show Solution
Graph y = ½x + 2: y-intercept (0,2), slope ½ (rise 1, run 2).
Graph y = 2x - 1: y-intercept (0,-1), slope 2 (rise 2, run 1).
The lines intersect at (2, 3).
Check: 3 = ½(2) + 2 = 3 ✓
3 = 2(2) - 1 = 3 ✓
Answer: (2,3)
Question 4 · Solving by Graphing
Solve the linear system by graphing.
$$x + y = 4$$
$$2x - y = -1$$
What is the point of intersection?
Show Solution
Rewrite each in slope-intercept form.
x + y = 4 → y = -x + 4: y-intercept (0,4), slope -1.
2x - y = -1 → y = 2x + 1: y-intercept (0,1), slope 2.
The lines intersect at (1, 3).
Check: 1 + 3 = 4 ✓
2(1) - 3 = -1 ✓
Answer: (1,3)
Question 5 · Solving by Elimination
Solve the following linear system using elimination:
$$5x + 2y = -11$$
$$3x + 2y = -9$$
Show Solution
Subtract equation 2 from equation 1:
(5x + 2y) - (3x + 2y) = -11 - (-9)
2x = -2
x = -1
Then 3(-1) + 2y = -9 → 2y = -6 → y = -3
Solution: (-1, -3)
Answer: (-1,-3)
Question 6 · Solving by Elimination
Solve the following linear system using elimination:
$$7x + 3y = -17$$
$$6x + 2y = -14$$
Show Solution
Multiply equation 1 by 2: 14x + 6y = -34
Multiply equation 2 by 3: 18x + 6y = -42
Subtract: 4x = -8 → x = -2
Then 7(-2) + 3y = -17 → 3y = -3 → y = -1
Solution: (-2, -1)
Answer: (-2,-1)
Question 7 · Solving by Elimination
Solve the following linear system using elimination:
$$5x + 7y = 3$$
$$2x + 3y = 1$$
Show Solution
Multiply equation 1 by 2: 10x + 14y = 6
Multiply equation 2 by 5: 10x + 15y = 5
Subtract: -y = 1 → y = -1
Then 2x + 3(-1) = 1 → 2x = 4 → x = 2
Solution: (2, -1)
Answer: (2,-1)
Question 8 · Solving by Substitution
Solve the following linear system using substitution:
$$y = 2x - 3$$
$$x + y = 6$$
Show Solution
Substitute y = 2x - 3 into x + y = 6:
x + (2x - 3) = 6
3x - 3 = 6
3x = 9
x = 3
Then y = 2(3) - 3 = 3
Solution: (3, 3)
Answer: (3,3)
Question 9 · Solving by Substitution
Solve the following linear system using substitution:
$$2x + y = 6$$
$$3x + 2y = 10$$
Show Solution
From equation 1: y = 6 - 2x
Substitute into equation 2: 3x + 2(6 - 2x) = 10
3x + 12 - 4x = 10
-x = -2
x = 2
Then y = 6 - 2(2) = 2
Solution: (2, 2)
Answer: (2,2)
Question 10 · Solving by Substitution
Solve the following linear system using substitution:
$$4x + y = 0$$
$$x + 2y + 1 = 0$$
Show Solution
From equation 1: y = -4x
Substitute into equation 2: x + 2(-4x) + 1 = 0
x - 8x + 1 = 0
-7x = -1
x = 1/7
Then y = -4(1/7) = -4/7
Solution: (1/7, -4/7)
Answer: (1/7,-4/7)
Question 11 · Linear Systems Word Problems
Three soccer balls and a basketball cost $155. Two soccer balls and three basketballs cost $220. Find the cost of each ball.
Show Solution
Let s be the cost of a soccer ball and b be the cost of a basketball.
Formula/model:
3s + b = 155
2s + 3b = 220
Solve:
From the first equation, b = 155 - 3s.
Substitute into the second equation: 2s + 3(155 - 3s) = 220 → -7s = -245 → s = 35.
Then b = 155 - 3(35) = 50.
Therefore, soccer=$35, basketball=$50
Answer: soccer=$35, basketball=$50
Question 12 · Linear Systems Word Problems
The Sports Shop sells Adidas running shoes for $82 a pair and Air Jensen basketball shoes for $95 a pair. One day, the shop sells a combined 75 pairs totaling $6241 in sales. How many pairs of each shoe were sold?
Show Solution
Let a be the number of Adidas pairs and j be the number of Jensen pairs.
Formula/model:
a + j = 75
82a + 95j = 6241
Solve:
From the first equation, a = 75 - j.
Substitute into the second equation: 82(75 - j) + 95j = 6241 → 13j = 91 → j = 7.
Then a = 75 - 7 = 68.
Therefore, Adidas=68, Jensen=7
Answer: Adidas=68, Jensen=7
Question 13 · Linear Systems Word Problems
A blue spruce tree grows an average of 15 cm per year. An eastern hemlock grows an average of 10 cm per year. When planted, the blue spruce was 120 cm tall and the eastern hemlock was 180 cm tall. How many years after planting will the trees reach the same height?
Show Solution
Let y be the number of years after planting.
Formula/model:
120 + 15y = 180 + 10y
Solve:
5y = 60 → y = 12.
The height is 120 + 15(12) = 300.
Therefore, 12 years, 300 cm
Answer: 12 years, 300 cm
Common mistakes
- Sign slip when distributing a negative.
- Forgetting to multiply every term when scaling for elimination.
- Not back-substituting to find the second variable.
- Leaving a word answer as (x, y) without saying what x and y are.