Math 20-1
ID: B
Name: ________________________
M20-1: Systems of Equations Practice Test
1. Solve by graphing: *A sketch of your graph is required for
full marks* [2]
3
y = x −4
2
3
2
y = (x ) − 1
16
3. Algebraically solve: [2]
y = −4x +12
2
y = −2 (x − 2 ) + 2
.
2. Solve by graphing: *A sketch of your graph is required for
full marks* [2]
2x 2 + 16x + y + 27 = 0
4x + y + 11 = 0
.
4. Algebraically solve: [2]
16x 2 − 128x + 25y + 81 = 0
y+9= 0
.
.
1
Name: ________________________
ID: B
5. Algebraically solve the following system of
equations: [2]
y = −5x 2 + 9
7. Solve: [2]
−x 2 − x − y + 2 = 0
2x 2 + 4x + 3y − 4 = 0
y = 10x 2 − 4x − 10
.
.
.
6. Algebraically solve the following equation:
9x 2 − 8x = −4x 2 − 7x
[2]
.
.
2
ID: B
M20-1: Systems of Equations Practice Test
Answer Section
1. Solution: ÊÁË 4,2 ˆ˜¯
2. P1(–2,–3)
P2(–4,5)
3. Solution: ÁÊË 3,0 ˜ˆ¯
4. P1(–1,–9)
P2(9,–9)
5. When combined the equation reduces to: 15x 2 − 4x − 19 = 0
19
44
Thus: x = −1, y = 4 or x =
, y=
15
45
6. When combined the equation reduces to: −13x 2 + x = 0
1
Thus: x =
or 0
13
7. ÁÊË 2, − 4 ˜ˆ¯ , ÁÊË −1, 2 ˜ˆ¯
1
Math 20-1
ID: C
Name: ________________________
M20-1: Systems of Equations Practice Test
1. Solve by graphing: *A sketch of your graph is required for
full marks* [2]
4
y =− x
5
2
2
y = (x ) + 2
25
3. Algebraically solve: [2]
y = 4x+13
2
y = −2 (x + 3 ) − 1
.
2. Solve by graphing: *A sketch of your graph is required for
full marks* [2]
3x 2 + 24x + 4y + 52 = 0
3x + 2y + 14 = 0
.
4. Algebraically solve: [2]
x 2 − 12x + 3y + 33 = 0
y+2= 0
.
.
1
Name: ________________________
ID: C
5. Algebraically solve the following system of
equations: [2]
y = 8x 2 − 9x
y = 5x − 5
7. Solve: [2]
x 2 − 4x + 2y + 10 = 0
x2 − x − y − 5 = 0
.
.
.
6. Algebraically solve the following equation:
11x 2 − 10x + 10 = 11x 2 − 9x + 8
[2]
.
.
2
ID: C
M20-1: Systems of Equations Practice Test
Answer Section
1. Solution: ÊÁË −5,4 ˆ˜¯
2. P1(–2,–4)
P2(–4,–1)
3. Solution: ÁÊË −4,−3 ˜ˆ¯
4. P1(9,–2)
P2(3,–2)
5. When combined the equation reduces to: −8x 2 + 14x − 5 = 0
5
5
1
5
Thus: x = , y = or x = , y = −
4
4
2
2
6. When combined the equation reduces to: x − 2 = 0
Thus: x = 2
7. ÊÁË 0, − 5 ˆ˜¯ , ÊÁË 2, − 3 ˆ˜¯
1