Simultaneous Equations: Line Meets a Curve

One linear plus one quadratic equation is the standard “line meets curve” pair. The method the mark scheme expects is fixed: substitute the linear equation into the quadratic, solve the resulting quadratic, then pair each answer back through the linear equation.

Solve the simultaneous equations $y = x^2 - 2x + 1 \qquad \text{and} \qquad y = x + 1.$ [5]

The working

Step 1, set the two expressions for $y$ equal (both equal $y$, so equate them): $x^2 - 2x + 1 = x + 1 \quad \text{(M1)}$

Step 2, collect to one side to form a quadratic $= 0$: $x^2 - 2x + 1 - x - 1 = 0 \;\Rightarrow\; x^2 - 3x = 0 \quad \text{(A1)}$

Step 3, factorise and solve (a common factor of $x$, no constant term): $x(x - 3) = 0 \;\Rightarrow\; x = 0 \quad \text{or} \quad x = 3 \quad \text{(M1)}$

Step 4, find each $y$ from the linear equation $y = x + 1$: $x = 0 \Rightarrow y = 1, \qquad x = 3 \Rightarrow y = 4$

Solutions: $(0, 1)$ and $(3, 4)$. (A1, A1)

Where the marks are won and lost

• Equating the two right-hand sides is valid because both equal $y$. Otherwise, substitute the linear form into the quadratic.
• $x^2 - 3x = 0$ factors to $x(x - 3)$, do not divide both sides by $x$ (that throws away the root $x = 0$).
• Pair each $x$ with its own $y$ using the linear equation, and present the answers as coordinate pairs.

Common mistakes

• Cancelling the $x$ and losing the $x = 0$ solution.
• Solving for $x$ but forgetting the $y$-values.
• Sign slips when collecting $-2x - x = -3x$.

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