You met both methods at GCSE. At A Level they must be quick and completely reliable, because they turn up inside coordinate geometry, vectors, mechanics and partial fractions.
| Method | How it works | Best when… |
|---|---|---|
| Elimination | Scale the equations so one variable has matching coefficients, then add or subtract. | Both equations are in the form $ax + by = c$. |
| Substitution | Make one variable the subject of one equation and substitute into the other. | One equation is already $y = \ldots$ or $x = \ldots$, or one equation is non-linear. |
Solve $3x + 2y = 11$ and $5x - 3y = 12$.
Solve $y = 2x - 5$ and $4x + 3y = 25$.
| Two straight lines | Number of solutions | Algebraic sign |
|---|---|---|
| Different gradients — they cross | Exactly one | Elimination gives a single value |
| Parallel and distinct | None | Elimination gives a contradiction such as $0 = 1$ |
| The same line | Infinitely many | Elimination gives $0 = 0$ |
Show that $2x + 4y = 7$ and $x + 2y = 3$ have no solution, and interpret this geometrically.
The same idea extends to curves. A line can meet a parabola or a circle in two points, one point (where it is a tangent) or none.
Substitute into the quadratic · solve the resulting quadratic · find the matching values
Solve the simultaneous equations $y = x + 1$ and $x^2 + y^2 = 25$.
Solve $x - 2y = 1$ and $x^2 + 2y^2 = 11$.
Find the coordinates of the points where the line $y = 2x - 3$ meets the curve $y = x^2 - 4x + 5$.
After substituting, you are left with a quadratic. Its discriminant tells you how many times the line meets the curve — without solving anything.
$b^2 - 4ac = 0$: the line is a tangent (touches at one point)
$b^2 - 4ac < 0$: the line misses the curve
The line $y = 3x + c$ is a tangent to the curve $y = x^2 + x + 2$. Find $c$ and the point of contact.
Show that the line $y = x - 3$ does not meet the circle $x^2 + y^2 = 4$.
Find the set of values of $m$ for which the line $y = mx + 1$ meets the curve $y = x^2 + 2$ at two distinct points.
Elimination
Match coefficients; same signs subtract, different signs add.
Substitution
Subject of one, substitute into the other.
Linear + quadratic
Always substitute the linear one into the quadratic.
Pairing up
Find each $y$ from the linear equation.
Geometry
Solutions = points of intersection.
Parallel lines
No solution: elimination gives $0 = k \neq 0$.
Two points
Combined quadratic has $b^2 - 4ac > 0$.
Tangent
$b^2 - 4ac = 0$: one repeated root.
Miss
$b^2 - 4ac < 0$: no real roots.
Check
Substitute your answers into both original equations.
Solve $2x + 3y = 12$ and $5x - 2y = 11$.
▶ Show solution
Multiply by $2$ and $3$: $4x + 6y = 24$ and $15x - 6y = 33$.
Add: $19x = 57$, so $x = 3$.
$6 + 3y = 12$, so $y = 2$.
Solve $y = 3x - 1$ and $2x + 5y = 46$.
▶ Show solution
$2x + 5(3x - 1) = 46 \;\Rightarrow\; 17x - 5 = 46 \;\Rightarrow\; x = 3$
$y = 3(3) - 1 = 8$
Find the coordinates of the points of intersection of $y = x + 2$ and $y = x^2 - 4$.
▶ Show solution
$x^2 - 4 = x + 2 \;\Rightarrow\; x^2 - x - 6 = 0 \;\Rightarrow\; (x-3)(x+2) = 0$
$x = 3 \Rightarrow y = 5$; $x = -2 \Rightarrow y = 0$.
Points $(3,\,5)$ and $(-2,\,0)$.
Solve $x + y = 4$ and $x^2 + y^2 = 10$.
▶ Show solution
$y = 4 - x$: $x^2 + (4 - x)^2 = 10 \;\Rightarrow\; 2x^2 - 8x + 16 = 10$
$2x^2 - 8x + 6 = 0 \;\Rightarrow\; x^2 - 4x + 3 = 0 \;\Rightarrow\; (x-1)(x-3) = 0$
$x = 1, y = 3$ or $x = 3, y = 1$.
Solve $2x - y = 3$ and $xy = 2$.
▶ Show solution
$y = 2x - 3$: $x(2x - 3) = 2 \;\Rightarrow\; 2x^2 - 3x - 2 = 0$
$(2x + 1)(x - 2) = 0$, so $x = 2$ or $x = -\tfrac12$.
$x = 2 \Rightarrow y = 1$; $x = -\tfrac12 \Rightarrow y = -1 - 3 = -4$.
Check: $2 \times 1 = 2$ ✓ and $-\tfrac12 \times -4 = 2$ ✓.
Explain, with reference to their graphs, why the equations $3x - y = 2$ and $6x - 2y = 4$ have infinitely many solutions.
▶ Show solution
The second equation is exactly twice the first, so both rearrange to $y = 3x - 2$.
They describe the same line, so every point on that line satisfies both equations: there are infinitely many solutions. (Elimination gives $0 = 0$.)
Show that the line $y = 2x + 3$ is a tangent to the curve $y = x^2 + 4x + 4$, and find the point of contact.
▶ Show solution
$x^2 + 4x + 4 = 2x + 3 \;\Rightarrow\; x^2 + 2x + 1 = 0$
Discriminant: $2^2 - 4(1)(1) = 0$, so there is exactly one (repeated) root: the line is a tangent.
$(x + 1)^2 = 0 \Rightarrow x = -1$, and $y = 2(-1) + 3 = 1$. Point of contact $(-1,\, 1)$.
The line $y = kx - 2$ is a tangent to the curve $y = x^2 + 2x + 2$. Find the two possible values of $k$ and the corresponding points of contact.
▶ Show solution
$x^2 + 2x + 2 = kx - 2 \;\Rightarrow\; x^2 + (2 - k)x + 4 = 0$
Tangent: $(2 - k)^2 - 16 = 0$, so $2 - k = \pm 4$, giving $k = -2$ or $k = 6$.
$k = 6$: $x^2 - 4x + 4 = (x-2)^2 = 0$, so $x = 2$, $y = 10$. Point $(2,\,10)$.
$k = -2$: $x^2 + 4x + 4 = (x+2)^2 = 0$, so $x = -2$, $y = 2$. Point $(-2,\,2)$.
The line $x + y = k$ and the circle $x^2 + y^2 = 8$ are drawn on the same axes.
(a) Find the values of $k$ for which the line is a tangent to the circle, and the points of contact. (b) Find the set of values of $k$ for which the line meets the circle at two distinct points.
▶ Show solution
Substitute $y = k - x$: $x^2 + (k - x)^2 = 8 \;\Rightarrow\; 2x^2 - 2kx + k^2 - 8 = 0$.
Discriminant $= 4k^2 - 4(2)(k^2 - 8) = 64 - 4k^2$.
(a) Tangent: $64 - 4k^2 = 0$, so $k = \pm 4$.
$k = 4$: $2x^2 - 8x + 8 = 0 \Rightarrow (x - 2)^2 = 0$, point $(2,\,2)$. $k = -4$: point $(-2,\,-2)$.
(b) $64 - 4k^2 > 0 \Rightarrow k^2 < 16 \Rightarrow -4 < k < 4$.
A rectangle has sides $x$ cm and $y$ cm, a perimeter of $34$ cm and a diagonal of length $13$ cm.
(a) Write down two equations in $x$ and $y$. (b) Solve them to find the dimensions of the rectangle. (c) Show that no rectangle with perimeter $34$ cm can have a diagonal of length $12$ cm.
▶ Show solution
(a) Perimeter: $2x + 2y = 34$, so $x + y = 17$. Pythagoras: $x^2 + y^2 = 169$.
(b) $y = 17 - x$: $x^2 + (17 - x)^2 = 169 \;\Rightarrow\; 2x^2 - 34x + 289 = 169$
$2x^2 - 34x + 120 = 0 \;\Rightarrow\; x^2 - 17x + 60 = 0 \;\Rightarrow\; (x - 5)(x - 12) = 0$
$x = 5, y = 12$ or $x = 12, y = 5$: the rectangle is $5$ cm by $12$ cm.
(c) Now $x^2 + (17 - x)^2 = 144 \;\Rightarrow\; 2x^2 - 34x + 145 = 0$.
Discriminant $= 34^2 - 4(2)(145) = 1156 - 1160 = -4 < 0$.
There are no real solutions, so no such rectangle exists. (Geometrically, the line $x + y = 17$ misses the circle $x^2 + y^2 = 144$; the shortest possible diagonal is $\sqrt{144.5} \approx 12.02$ cm, for a square.)