IP AMaths Notes (Upper Sec, Year 3-4): 04) Simultaneous Equations

Expect to pair a linear relation with a curve or two curves together. Sketch the solution region mentally before diving into algebra so you keep track of feasible roots.

Toolkit

• Linear pair: eliminate a variable via addition/subtraction or substitution.
• Linear with quadratic: substitute the linear expression into the quadratic to obtain a single-variable equation, then solve and back-substitute.
• Quadratic pair: subtract to eliminate squared terms, or complete the square to interpret geometrically.

Worked example 1 — Linear and quadratic

Solve the system \( 2x + y = 7 \) and \( x^{2} + xy = 12 \).

1. Express \( y = 7 - 2x \).
2. Substitute: \( x^{2} + x(7 - 2x) = 12 \).
3. Simplify: \( x^{2} + 7x - 2x^{2} - 12 = 0 \) giving \( -x^{2} + 7x - 12 = 0 \).
4. Multiply by \( -1 \): \( x^{2} - 7x + 12 = 0 \).
5. Factor: \( (x - 3)(x - 4) = 0 \) so \( x = 3 \) or \( x = 4 \).
6. Back-substitute: when \( x = 3 \), \( y = 7 - 6 = 1 \); when \( x = 4 \), \( y = 7 - 8 = -1 \).

Solutions: \( (3, 1) \) and \( (4, -1) \).

Worked example 2 — Two quadratics

Solve \( x^{2} + y^{2} = 25 \) and \( y = x + 1 \).

1. Substitute \( y \): \( x^{2} + (x + 1)^{2} = 25 \).
2. Expand: \( x^{2} + x^{2} + 2x + 1 = 25 \) leading to \( 2x^{2} + 2x - 24 = 0 \).
3. Divide by \( 2 \): \( x^{2} + x - 12 = 0 \).
4. Factor: \( (x + 4)(x - 3) = 0 \).
5. Thus \( x = -4 \) or \( x = 3 \).
6. Corresponding \( y \): \( y = -3 \) or \( y = 4 \).

Solutions: \( (-4, -3) \) and \( (3, 4) \).

Try this

Solve \( xy = 6 \) and \( x + y = 5 \). Interpret the solutions on the \( xy \)-plane.