Edexcel 4MA1 only pairs one linear equation with one quadratic/non-linear equation (never two quadratics), so there is always a guaranteed route: substitute the linear equation into the quadratic one. This always collapses to a single quadratic in one variable, which you already know how to solve.

5-step method:

1. Isolate a variable from the linear equation — choose whichever avoids fractions (e.g. from $x - 2y = 1$, isolate $x = 1+2y$ rather than $y$, which gives a fraction).
2. Substitute that expression into the quadratic equation.
3. Expand and simplify into $aX^2+bX+c=0$, then solve by factorising or the quadratic formula.
4. Back-substitute both solutions into the simple linear equation to find their partner values.
5. Pair your answers clearly: $(x_1,y_1)$ and $(x_2,y_2)$ — never list $x$-values and $y$-values separately.

Expanding a squared bracket like $(9-x)^2$ as $81-x^2$ instead of $81-18x+x^2$. Always write it out twice: $(9-x)(9-x)$, and expand fully with FOIL. A close second is writing answers unpaired (e.g. "$x=3,4.5$ and $y=4.5,3$") — pair them explicitly.

The solution(s) are exactly where the two graphs intersect. A line meeting a circle can cross it twice (secant), touch it once (tangent), or miss it (no real solutions) — matching two, one, or zero algebraic solutions. Sketching the two graphs mentally before you start algebra is a powerful sense-check: if your answer pairs don't land where the sketch suggests, you've made an arithmetic slip.

Once substitution gives $aX^2+bX+c=0$, its discriminant $\Delta = b^2-4ac$ reveals the number of intersection points without solving fully:

• $\Delta > 0$: two distinct solutions (line is a secant)
• $\Delta = 0$: one repeated solution (line is a tangent)
• $\Delta < 0$: no real solutions (line misses the curve)

This is how "show the line is a tangent to the curve" questions are answered: substitute, form the quadratic, and show $\Delta = 0$.

Grade 9 questions often don't give you the equations — you must build them. Perimeter/sum-type statements are linear; area/product-type statements are quadratic (note: $xy$ is degree 2, just like $x^2$).

Worked example: A rectangle has perimeter 30 cm and area 54 cm², with length greater than width. Find its dimensions.

Let length $= l$, width $= w$.

Substitute $l = 15-w$:

So $w=6$ or $w=9$. Since $l=15-w$: pairs are $(l,w)=(9,6)$ or $(6,9)$. As $l>w$, the length is 9 cm and width is 6 cm.

Solve algebraically: $y = 9-x$ and $x^2-3xy+2y^2=0$.

Substitute $y=9-x$:

Expand each term carefully: $-3x(9-x) = -27x+3x^2$, and $2(9-x)^2 = 2(81-18x+x^2) = 162-36x+2x^2$.

Divide by 3: $2x^2-21x+54=0$. Using the quadratic formula:

Using $y=9-x$: if $x=6$, $y=3$; if $x=4.5$, $y=4.5$.

Solutions: $(6,3)$ and $(4.5,4.5)$.