Solve two simultaneous linear equations by elimination and substitution; solve a linear and a quadratic equation simultaneously (Higher tier); and interpret the solution as the point of intersection.

What this dot point is asking

The Eduqas algebra content asks you to solve a pair of simultaneous equations: two equations with two unknowns that hold at the same time. You must solve two linear equations by elimination and by substitution, and at Higher tier solve one linear and one quadratic equation together. You should also interpret the solution geometrically as the point where the two graphs intersect. Simultaneous equations appear on both components and reward clear, organised working, and the linear-quadratic case is a reliable Higher-tier multi-mark question.

Solving by elimination

Elimination removes one unknown so a single equation in the other can be solved. Line the equations up, then add or subtract.

For $2x + 3y = 13$ and $4x - 3y = 5$, the $y$ terms are $+3y$ and $-3y$, so adding gives $6x = 18$ and $x = 3$. Substituting back into the first equation, $6 + 3y = 13$, so $y = \tfrac{7}{3}$.

When no coefficients match, scale first. For $3x + 2y = 12$ and $4x + y = 11$, multiply the second by $2$ to get $8x + 2y = 22$, then subtract the first to eliminate $y$: $5x = 10$, so $x = 2$.

Solving by substitution

Substitution rearranges one equation to make a variable the subject, then puts that expression into the other equation. It is the natural choice when one equation already gives a variable explicitly, such as $y = 3x - 1$.

A linear and a quadratic (Higher)

At Higher tier one equation is a quadratic and the other is linear. Substitution is the only method that works: make a variable the subject of the linear equation, substitute into the quadratic, and solve the single quadratic that results.

For $y = x^2 + 1$ and $y = x + 3$, set them equal: $x^2 + 1 = x + 3$, which rearranges to $x^2 - x - 2 = 0$. This factorises as $(x - 2)(x + 1) = 0$, so $x = 2$ or $x = -1$. The matching $y$ values from $y = x + 3$ are $5$ and $2$, giving the two intersection points $(2, 5)$ and $(-1, 2)$. Always pair each $x$ with its own $y$; reporting them mismatched loses marks.

Forming a pair from a context

Many Eduqas problems hide a simultaneous pair inside a word problem. Two adults and three children pay 23 pounds, while one adult and four children pay 19 pounds: with $a$ the adult price and $c$ the child price, this becomes $2a + 3c = 23$ and $a + 4c = 19$. Solving (multiply the second by $2$, then subtract) gives $c = 3$ and $a = 7$. The skill Eduqas rewards is the translation: choose a letter for each unknown, write one equation per piece of information, then solve by the method that fits. Always define your variables in words first so the examiner can follow the reasoning.

The point of intersection

Each linear equation is a straight line and each quadratic is a curve. The simultaneous solution is exactly where the graphs cross, so two linear equations meet at one point (unless parallel), and a line and a parabola can meet at two points, one point (a tangent), or none. This geometric reading is why Eduqas may give a graph and ask you to "use the diagram to estimate the solution", then confirm it algebraically. It also explains the special cases: two parallel lines (equal gradients) never meet, so the equations have no solution, while two copies of the same line meet everywhere.