GRADE 9 MATHEMATICS
Learning objective 9.2.2.2
Teacher answer key
Four nonlinear-system methods | 20 problems
All solutions are real ordered pairs (x, y). For the graphical worksheet, intersection coordinates are rounded to two decimal
places when needed.
1. Substitution method
After expressing y in terms of x, substitute into the nonlinear equation and factor.
1. y = 7 − x, so 2(x − 3)(x − 4) = 0.
Solutions: (3, 4), (4, 3)
2. y = 3 − x, so 3x(x − 4) = 0.
Solutions: (0, 3), (4, −1)
3. y = 1 − x, so 3(x − 2)(x + 1) = 0.
Solutions: (−1, 2), (2, −1)
4. y = x − 2, so 2(x − 2)(x + 3) = 0.
Solutions: (−3, −5), (2, 0)
5. y = 2 − x, so (x − 1)(4x + 7) = 0.
7 15
, (1, 1)
Solutions: − ,
4 4
Marking check: Award evidence for the expressed variable, substitution, correct quadratic, roots, and complete ordered pairs.
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GRADE 9 MATHEMATICS
Learning objective 9.2.2.2
2. Change of variables
√
For each question use u = A2 and v = B 2 , with A, B shown below. Solve the resulting linear system and use A = ± u and
√
B = ± v.
1. A = x − 1, B = y + 1; (u, v) = (1, 4).
Solutions: (0, −3), (0, 1), (2, −3), (2, 1)
2. A = x + y, B = x − y; (u, v) = (16, 4).
Solutions: (−3, −1), (−1, −3), (1, 3), (3, 1)
3. A = x − 2, B = x − 2y; (u, v) = (9, 4).
1
3
7
3
, −1,
, 5,
, 5,
Solutions: −1, −
2
2
2
2
4. A = x + y, B = 2x − y; (u, v) = (9, 4).
5
4
1
8
1 8
5 4
Solutions: − , −
, − ,−
,
,
,
,
3
3
3
3
3 3
3 3
5. A = x − 3y, B = x + y; (u, v) = (4, 9).
1
7
5
7 5
11 1
11
, − ,−
,
,
,
,
Solutions: − , −
4
4
4
4
4 4
4 4
Marking check: Check all sign choices and both substitutions; four real ordered pairs are expected in each question.
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GRADE 9 MATHEMATICS
Learning objective 9.2.2.2
3. Algebraic transformations
The right-hand sides are nonzero, so the shared factors that would make both left-hand sides zero cannot vanish for a solution.
1. x(x + y) = 6, y(x + y) = 3; divide to get x = 2y, then 3y 2 = 3.
Solutions: (−2, −1), (2, 1)
2. x(x + y) = 12, y(x + y) = 4; x = 3y, then 4y 2 = 4.
Solutions: (−3, −1), (3, 1)
3. x(x − y) = 8, y(x − y) = 4; x = 2y, then y 2 = 4.
Solutions: (−4, −2), (4, 2)
4. 2x(x + y) = 30, 3y(x + y) = 30; x(x + y) = 15, y(x + y) = 10; x = 32 y, 25 y 2 = 10.
Solutions: (−3, −2), (3, 2)
5. x(x + y) = 12, y(x + y) = −3; x = −4y, then −3y 2 = −3.
Solutions: (−4, 1), (4, −1)
Marking check: The division step is valid only after noting the denominator/nonzero common factor.
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GRADE 9 MATHEMATICS
Learning objective 9.2.2.2
4. Graphical method
Complete the square, graph both curves, and read off each common point.
1. (x − 1)2 + (y − 1)2 = 5, y = 4 − (x − 1)2 .
Solutions: (−1, 0), (0, 3), (2, 3), (3, 0)
2. (x + 1)2 + (y − 2)2 = 5, y = 5 − (x + 1)2 .
Solutions: (−3, 1), (−2, 4), (0, 4), (1, 1)
3. (x − 2)2 + (y + 1)2 = 5, y = 2 − (x − 2)2 .
Solutions: (0, −2), (1, 1), (3, 1), (4, −2)
4. (x + 1)2 + (y − 1)2 = 25, y = 6 − (x + 1)2 .
Solutions: (−4, −3), (−1, 6), (2, −3)
5. (x − 2)2 + y 2 = 9, y = 4 − x2 .
Solutions: (1.07, 2.85), (2.63, −2.93)
Marking check: Correct transformed equations, visible circle and parabola, and all intersection points. Question 5 has
approximate graph-read answers.
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