Simultaneous Equations with an xy Product

A product equation such as $xy = 20$ is handled the same way as any non-linear pair: substitute the linear equation in, expand, and solve the resulting quadratic. The factorising step, with a non-unit leading coefficient, is where the care goes.

Solve the simultaneous equations $2x - y = 3 \qquad \text{and} \qquad xy = 20.$ [5]

The working

Step 1, make $y$ the subject of the linear equation: $y = 2x - 3 \quad \text{(M1)}$

Step 2, substitute into $xy = 20$: $x(2x - 3) = 20 \;\Rightarrow\; 2x^2 - 3x = 20 \quad \text{(M1)}$

Step 3, form the quadratic $= 0$ and factorise: $2x^2 - 3x - 20 = 0 \;\Rightarrow\; (2x + 5)(x - 4) = 0 \quad \text{(A1)}$

Step 4, solve and pair via $y = 2x - 3$: $x = 4 \Rightarrow y = 5, \qquad x = -\frac{5}{2} \Rightarrow y = 2\!\left(-\tfrac{5}{2}\right) - 3 = -8 \quad \text{(A1, A1)}$

Solutions $(4, 5)$ and $\left(-\frac{5}{2}, -8\right)$.

Where the marks are won and lost

• Distributing $x(2x - 3) = 2x^2 - 3x$ correctly is the second method mark, a sign slip here derails the quadratic.
• Factorising $2x^2 - 3x - 20$: the split is $(2x + 5)(x - 4)$ (check by expanding: $2x^2 - 8x + 5x - 20 = 2x^2 - 3x - 20$ ✓). The quadratic formula is a safe fallback.
• The negative solution is valid here (nothing restricts $x$), so both pairs are required.

Common mistakes

• Expanding $x(2x - 3)$ as $2x^2 - 3$ (dropping the $x$).
• Factorising errors on the non-unit leading coefficient.
• Discarding the fractional/negative solution without reason.

Full method: One Linear + One Non-Linear notes. Topic home: Simultaneous Equations pillar.