Line up both equations, then add or subtract them to cancel one variable:

• Same sign, same size (e.g. both $+3y$) → subtract.
• Opposite sign, same size (e.g. $+2y$ and $-2y$) → add.
• If the coefficients don't match, multiply one or both equations first.

Example: $3x+y=5$ and $x+y=1$. Both have $+y$, so subtract: $2x=4 \Rightarrow x=2$. Substitute back: $2+y=1 \Rightarrow y=-1$.

Rearrange one equation to isolate a variable, then substitute that expression into the other equation.

Example: $y=2x+1$ and $5x+3y=14$. Substitute: $5x+3(2x+1)=14 \Rightarrow 11x+3=14 \Rightarrow x=1$, then $y=3$.

Substitution is the natural choice whenever one equation is already arranged as $y=\ldots$, and it's the only method that works once a non-linear equation (like a quadratic) is involved.

Elimination cannot handle a squared term, so substitution is mandatory:

1. Rearrange the linear equation to isolate a variable.
2. Substitute into the quadratic equation.
3. Rearrange into $Ax^2+Bx+C=0$ and solve (two $x$-values are typical).
4. Substitute each $x$-value back into the simple linear equation to find its paired $y$-value.
5. State both solutions as coordinate pairs.

Two solutions make geometric sense: a straight line can cross a parabola at up to two points.

Worked Example

Solve $y=2x-1$ and $y=x^2-4$ simultaneously.

1. Set equal: $2x-1=x^2-4$.
2. Rearrange: $x^2-2x-3=0$.
3. Factorise: $(x+1)(x-3)=0 \Rightarrow x=-1$ or $x=3$.
4. Find $y$: for $x=-1$, $y=2(-1)-1=-3$; for $x=3$, $y=2(3)-1=5$.

Solutions: $(-1,-3)$ and $(3,5)$.

Examiners consistently flag two issues: using trial-and-improvement instead of a proper algebraic method (which scores no method marks even with a correct answer), and losing track of which $x$ pairs with which $y$ when there are two solutions. Always label your equations (1) and (2), and write out the elimination/substitution operation explicitly (e.g. "(1) − (2)") rather than doing it in your head.

Elimination only works while every equation stays linear. The moment a quadratic term appears, elimination has no way to cancel it out, and substitution becomes the only algebraic route to a solution. Building confidence with substitution on simple linear pairs first means the jump to linear/quadratic systems is just an extra couple of steps (expanding and solving a quadratic), not an entirely new technique learned under exam pressure.

Substitute your final values back into both original equations — not just the one you used to find the second variable. This catches errors made during rearrangement or expansion that wouldn't otherwise be obvious, and for linear/quadratic systems, it also confirms you've paired each $x$ with its correct $y$.