Solving linear simultaneous equations by elimination and substitution, and solving a linear and quadratic pair at Higher tier.

What this dot point is asking

AQA wants you to solve a pair of simultaneous equations, by elimination or substitution for two linear equations, and at Higher tier by substitution for a linear-and-quadratic pair. The two equations together pin down a single point (or points) that satisfy both, which graphically is where the two lines or curves intersect. This skill is tested directly and also underpins problem-solving questions phrased in words.

Solving by elimination

Elimination is usually fastest for two linear equations. Make the coefficient of one variable equal in size, then add the equations (if the signs differ) or subtract them (if the signs match) to remove that variable.

When the coefficients do not match, scale one or both equations first. For $3x + 4y = 18$ and $2x + y = 7$, multiply the second equation by $4$ to get $8x + 4y = 28$, then subtract the first to eliminate $y$.

Solving by substitution

Substitution works well when one equation already has a variable on its own. From $y = 2x - 1$ and $3x + y = 9$, substitute the first into the second: $3x + (2x - 1) = 9$, so $5x - 1 = 9$, giving $x = 2$ and then $y = 3$.

Linear and quadratic pairs at Higher tier

When one equation is quadratic, substitution is the only viable method. Rearrange the linear equation to make one variable the subject, substitute into the quadratic, and you get a single quadratic equation to solve. Because a line can cut a curve twice, expect up to two solution pairs.

For $x^2 + y^2 = 25$ and $y = x + 1$, substitute: $x^2 + (x + 1)^2 = 25$, expand to $x^2 + x^2 + 2x + 1 = 25$, simplify to $2x^2 + 2x - 24 = 0$, divide by $2$ to get $x^2 + x - 12 = 0$, factorise to $(x + 4)(x - 3) = 0$, so $x = -4$ or $x = 3$. The matching $y$ values from $y = x + 1$ are $y = -3$ and $y = 4$, giving the points $(-4, -3)$ and $(3, 4)$.

Interpreting the solution graphically

The solution of a linear pair is the single point where the two lines cross. If the lines are parallel there is no solution; if they are identical there are infinitely many. For a linear-and-quadratic pair, the solutions are the intersection points of the line with the curve: two points, one (a tangent), or none.

Forming simultaneous equations from words

Many of the marks come from setting up the equations from a worded scenario. The standard structure gives two pieces of information about two unknowns. "Three coffees and two teas cost $\pounds 11$; one coffee and two teas cost $\pounds 7$" becomes $3c + 2t = 11$ and $c + 2t = 7$. Subtracting eliminates $t$ at once: $2c = 4$, so $c = 2$ and then $t = 2.5$. Defining the variables clearly ("let $c$ be the price of a coffee") at the start secures the setup marks even if the arithmetic later slips.

Choosing elimination or substitution

Both methods always reach the same answer, so choose by what is in front of you. Use elimination when both equations are in the tidy form $ax + by = c$, since matching a coefficient and adding or subtracting is quick. Use substitution when one equation already gives a variable on its own (such as $y = 3x - 2$) or when one equation is quadratic, where substitution is the only route. With practice you can spot the faster path immediately, but on equal terms elimination tends to involve less algebra for two linear equations.