← Back to the Systems Of Linear Equations worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: A-REI.C.5, A-REI.C.6, A-REI.D.11, A-CED.A.3 · Level: Grade 9 Algebra 1
Problem 1 Check a solution
First equation: 3 + 1 = 4 ✓. Second equation: 2(3) − 1 = 5 ✓.
Yes, (3, 1) is a solution.
Problem 2 Check a failing pair
First: 3(2) − 7 = −1 ✓. Second: 2 + 2(−1) = 0, not 1.
No, it fails the second equation.
Problem 3 Substitution with y = 2x
x + 3x = 28, so 4x = 28 and x = 7. Then y = 21.
Check: 7 + 21 = 28 ✓.
(7, 21)
Problem 4 Substitution with x = y + 4
2(y + 4) + y = 17, so 3y + 8 = 17 and y = 3. Then x = 7.
Check: 2(7) + 3 = 17 ✓.
(7, 3)
Problem 5 Read the graph
The lines cross where x = −1 and y = 2. Check with equations y = x + 3 and y = −2x: 2 = −1 + 3 ✓ and 2 = −2(−1) ✓.
(−1, 2)
Problem 6 Elimination by adding
Add: 2x = 20, so x = 10. Then 10 + y = 15 gives y = 5.
Check: 10 − 5 = 5 ✓.
(10, 5)
Problem 7 Elimination with opposite terms
Add: 11x = 33, so x = 3. Then 21 + 3y = 27 gives y = 2.
Check: 12 − 6 = 6 ✓.
(3, 2)
Problem 8 Scale one equation
Multiply the second equation by 5: 15x − 5y = 25. Add to the first: 17x = 51, so x = 3.
Then 3(3) − y = 5 gives y = 4.
Check: 2(3) + 5(4) = 26 ✓.
(3, 4)
Problem 9 Scale both equations
Multiply the first by 2 and the second by 3: 8x + 6y = 50 and 9x − 6y = 18. Add: 17x = 68, so x = 4.
Then 16 + 3y = 25 gives y = 3.
Check: 12 − 6 = 6 ✓.
(4, 3)
Problem 10 No solution
Subtract the equations: 0 = 6, which is false.
The graph shows parallel lines with the same slope −1/2 and different intercepts.
No solution
Problem 11 Infinitely many solutions
Multiply the second equation by 2: 10x − 4y = 14, identical to the first.
Subtracting gives 0 = 0, a true statement.
Infinitely many solutions (same line)
Problem 12 Why elimination works
Subtract 2(x + y) = 18 from 2x + 5y = 33: 3y = 15, so y = 5 and x = 4.
The original system gives 4 + 5 = 9 and 8 + 25 = 33, so (4, 5) satisfies both systems.
The new second equation is true whenever both originals are true, so the solution is unchanged.
(4, 5) in both systems
Problem 13 Solve by graphing
The first line has y-intercept −1 and slope 2, passing through (0, −1) and (3, 5); the second passes through (0, 8) and (8, 0). They cross at (3, 5).
Check: 2(3) − 1 = 5 ✓ and −3 + 8 = 5 ✓.
(3, 5)
Problem 14 Two functions equal
Set 4x − 7 = −x + 13, so 5x = 20 and x = 4.
f(4) = 9 and g(4) = −4 + 13 = 9.
The graphs of f and g cross at (4, 9).
x = 4 and the common output 9
Problem 15 Ticket sales
Let a be adults and s students: a + s = 480 and 12a + 7s = 4,530.
Substitute s = 480 − a: 12a + 7(480 − a) = 4,530, so 5a + 3,360 = 4,530 and a = 234. Then s = 246.
Check: 234(12) + 246(7) = 2,808 + 1,722 = 4,530 ✓.
234 adult and 246 student tickets
Problem 16 Coffee mixture
Let x and y be pounds: x + y = 10 and 8x + 14y = 104 (since 10 × 10.40 = 104).
8x + 14(10 − x) = 104 gives −6x = −36 and x = 6, so y = 4.
Check: 48 + 56 = 104 ✓.
6 pounds of $8 beans and 4 pounds of $14 beans
Problem 17 Phone plans
24 + 0.08m = 9 + 0.18m, so 15 = 0.10m and m = 150. Bill: 24 + 12 = $36.
At 150 minutes; $36
Problem 18 Kayak and river
Let b be the paddling speed and c the current. With the current: b + c = 18 ÷ 2 = 9. Against it: b − c = 18 ÷ 3 = 6.
Add: 2b = 15, so b = 7.5 and c = 1.5.
Kayaker 7.5 mi/h; river 1.5 mi/h
Problem 19 Nickels and dimes
n + d = 35 and 5n + 10d = 280 (in cents).
5n + 10(35 − n) = 280 gives −5n = −70, so n = 14 and d = 21.
Check: 70 + 210 = 280 ✓.
14 nickels and 21 dimes
Problem 20 Garden dimensions
l = 2w + 3 and 2l + 2w = 66.
2(2w + 3) + 2w = 66 gives 6w + 6 = 66 and w = 10, so l = 23.
Check: 2(23) + 2(10) = 66 ✓.
10 feet by 23 feet