systems of three equations: solve, and count the solutions
Grade 11 Math Worksheet

Circle the best answer.
Read this first
Use the table
Equations: (1) -3x + 2y + z = 12; (2) 3x − 4y − z = -4; (3) 6x + 5y + 2z = -40.
x y z total (1) -3 2 1 12 (2) 3 -4 -1 -4 (3) 6 5 2 -40 Luz cancelled z with rows (1) and (2). Row (3) has a z number twice the size of row (1)'s. Which step on rows (1) and (3) cancels z?
- Add rows (1) and (3)
- Subtract row (3) from row (1)
- Add 2 × row (1) to row (3)
- Subtract row (3) from 2 × row (1)
Noah cancelled z from rows (1) and (2) and now works on rows (1) and (3). Which step cancels z?
- Subtract row (3) from row (1)
- Subtract row (3) from 2 × row (1)
- Add rows (1) and (3)
- Add 2 × row (1) to row (3)
Pia multiplies row (1) by 2 and subtracts it from row (2) to cancel x. Which new row is left?
- 13y − 6z = 5
- 4x − 3y + 10z = -16
- -3y − 6z = -16
- 13y − 6z = 12
To cancel x, Ren subtracts 3 × row (1) from row (2) of the number table. Which new row is left?
- -8y − 7z = -46
- 16y − 7z = 50
- 6x − 8y + 5z = -46
- 16y − 7z = 18
In Ari's table, row (2) minus 2 × row (1) cancels x. Which new row is left?
- 9y + 3z = 60
- 4x − 3y − 13z = -72
- 9y + 3z = 27
- -3y − 13z = -72
Use the table
Equations: (1) 2x + 2y + z = -8; (2) -5x − 4y − z = 6; (3) 4x + 2y − z = 11.
x y z total (1) 2 2 1 -8 (2) -5 -4 -1 6 (3) 4 2 -1 11 Row (4) = row (1) + row (2): -3x − 2y = -2. Row (5) = row (1) + row (3): 6x + 4y = 3.
Bo adds row (1) to rows (2) and (3) to cancel z, giving rows (4) and (5). How many solutions does the system have?
- No solution
- Infinitely many solutions
- Exactly one solution
Use the table
Equations: (1) 2x + 3y − z = 3; (2) -6x − 4y + z = -10; (3) 2x − 2y + z = 4.
x y z total (1) 2 3 -1 3 (2) -6 -4 1 -10 (3) 2 -2 1 4 Row (4) = row (1) + row (2): -4x − y = -7. Row (5) = row (1) + row (3): 4x + y = 7.
Esme cancels z, making rows (4) and (5) from the table. How many solutions does the system have?
- Exactly one solution
- Infinitely many solutions
- No solution
Use the table
Equations: (1) 3x − 2y + z = -3; (2) -7x + 3y − z = 8; (3) -2x + 6y − z = -11.
x y z total (1) 3 -2 1 -3 (2) -7 3 -1 8 (3) -2 6 -1 -11 Row (4) = row (1) + row (2): -4x + y = 5. Row (5) = row (1) + row (3): x + 4y = -14.
Adding row (1) to rows (2) and (3) gives Jun rows (4) and (5) without z. How many solutions does the system have?
- Exactly one solution
- No solution
- Infinitely many solutions
Use the table
Equations: (1) -x − 2y + 3z = -5; (2) -x − 2y − 3z = 1; (3) -3x − 6y − 3z = -4.
x y z total (1) -1 -2 3 -5 (2) -1 -2 -3 1 (3) -3 -6 -3 -4 Row (4) = row (1) + row (2): -2x − 4y = -4. Row (5) = row (1) + row (3): -4x − 8y = -9.
Kira adds row (1) to rows (2) and (3) to cancel z, giving rows (4) and (5). How many solutions does the system have?
- Exactly one solution
- Infinitely many solutions
- No solution
Use the table
Equations: (1) x − y − 3z = -2; (2) 2x + 2y + 3z = 4; (3) 8x + 4y + 3z = 8.
x y z total (1) 1 -1 -3 -2 (2) 2 2 3 4 (3) 8 4 3 8 Row (4) = row (1) + row (2): 3x + y = 2. Row (5) = row (1) + row (3): 9x + 3y = 6.
Luz cancels z, making rows (4) and (5) from the table. How many solutions does the system have?
- No solution
- Exactly one solution
- Infinitely many solutions

