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Simultaneous equations and cubic equations
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When one equation is linear and one is quadratic, substitute. When a cubic has a rational root, the factor theorem cracks it open.
Linear and quadratic together
Rearrange the linear equation for x or y. Substitute into the quadratic. Solve, then find the matching values.
y = 2x − 1 and x2 + y2 = 10
x2 + (2x − 1)2 = 10
5x2 − 4x − 9 = 0
(5x − 9)(x + 1) = 0, so x = −1 or x = 1.8
y = 2x − 1 and x2 + y2 = 10
x2 + (2x − 1)2 = 10
5x2 − 4x − 9 = 0
(5x − 9)(x + 1) = 0, so x = −1 or x = 1.8
Pair the answers
x = −1 gives y = −3. x = 1.8 gives y = 2.6.
Solutions: (−1, −3) and (1.8, 2.6). Always use the linear equation to find the partner value.
Solutions: (−1, −3) and (1.8, 2.6). Always use the linear equation to find the partner value.
Solving cubics
x3 − 2x2 − 5x + 6 = 0
f(1) = 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor.
Divide: (x − 1)(x2 − x − 6) = (x − 1)(x − 3)(x + 2) = 0
x = 1, 3 or −2
f(1) = 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor.
Divide: (x − 1)(x2 − x − 6) = (x − 1)(x − 3)(x + 2) = 0
x = 1, 3 or −2
Solve x + y = 4 and xy = 3.
- y = 4 − x
- x(4 − x) = 3, so x2 − 4x + 3 = 0
- (x − 1)(x − 3) = 0
Answer: x = 1, y = 3 or x = 3, y = 1
Substitute the linear equation into the quadratic, solve, and pair each x with its y. For a cubic, find a root with the factor theorem, divide, then solve the quadratic.
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