Simultaneous Equations

A short topic with a single method, and a quiet importance well beyond its own marks, because line meets curve is the geometry behind discriminant questions, circle–line intersections and area-between-curves setups. Master the routine once and four topics pay out.

The method: substitute from the linear

Solve $y = x + 3$ and $x^2 + y^2 = 29$ simultaneously.

1. Linear is already $y = x + 3$, if not, rearrange it (never try to rearrange the quadratic).
2. Substitute into the non-linear: $x^2 + (x + 3)^2 = 29$, write this line with the brackets intact (it’s the method mark).
3. Expand and collect to zero: $x^2 + x^2 + 6x + 9 = 29 \to 2x^2 + 6x - 20 = 0 \to x^2 + 3x - 10 = 0$
4. Solve the quadratic: $(x + 5)(x - 2) = 0 \to x = -5$ or $x = 2$
5. Back-substitute into the LINEAR: $x = -5 \to y = -2$; $x = 2 \to y = 5$
6. Present as pairs: $(-5, -2)$ and $(2, 5)$

Steps 2, 3, 4 each carry method/accuracy credit; steps 5 to 6 carry the final marks, and step 5 done into the quadratic instead of the linear is this topic’s signature self-sabotage (it invites extra, invalid pairings and harder algebra).

Points of intersection: same algebra, geometric language

“Find the points where the line meets the curve” is the identical computation. The number of solutions tells the geometry: two = crosses, one (repeated) = tangent, none = misses. When the question asks only whether or for what $k$ the line meets the curve, no coordinates wanted, skip solving and go straight to the discriminant of the collected quadratic. Reading what the question actually wants (command words) chooses between the five-step solve and the one-line discriminant.

Pairing discipline

Each $x$ belongs to its own $y$. Writing ”$x = -5, 2$; $y = -2, 5$” without pairing can cost the final mark, and inviting the examiner to wonder whether $(-5, 5)$ was intended never ends well. Coordinates, as pairs, every time. If the two curves are needed later (say, as limits of an area integral), labelled pairs prevent downstream errors too.

Worked exam-style variant, the tangent twist

The line $y = 3x + c$ is a tangent to the curve $y = x^2 - x + 7$. Find $c$.

Substitute: $x^2 - x + 7 = 3x + c \to x^2 - 4x + (7 - c) = 0$ (M) Tangent $\Rightarrow$ equal roots $\Rightarrow b^2 - 4ac = 0$: $16 - 4(7 - c) = 0$ (M, discriminant stated) $16 - 28 + 4c = 0 \to \mathbf{c = 3}$ (A)

No solving for points at all, the discriminant shortcut is the intended method, and attempting the full solve wastes five minutes the paper doesn’t give you (timing strategy).

Common mistakes in this topic

• Substituting from the quadratic into the linear (creates square roots and chaos)
• Bracket dropped when substituting: $x + 3^2$ instead of $(x + 3)^2$
• Back-substitution into the wrong equation
• Solutions left unpaired, or paired across
• Solving fully when only the discriminant was needed, correct answer, lost time

On the non-calculator paper the quadratics here factorise cleanly by design; reaching for the formula and surds usually signals an expansion slip upstream.

One clean afternoon of drilling makes this topic automatic. If it still wobbles, bring it to the free 1-hour trial class, message Teacher Rig on WhatsApp.