Graphical solutions to simultaneous equations Revision Notes | GCSE AQA Higher Maths

1. Why Graphs Solve Equations

Here is a question worth sitting with for a moment: if two equations both describe relationships between $x$ and $y$, what does it mean for a point to satisfy both at once? That point has to live on both graphs simultaneously — which means it must be where the two graphs cross.

That is the entire idea behind graphical solutions to simultaneous equations. Instead of manipulating algebra, you draw both graphs on the same axes and read off the coordinates of any intersection points. Those coordinates are your solutions.

Graphical solutions to two straight lines are taught across both tiers, because reading the intersection of two linear graphs is standard GCSE content. The Higher-only extension in this area is the linear-quadratic case, where a straight line meets a curve and there may be two solutions, one tangent solution, or no real intersection. This lesson therefore focuses mainly on linear-quadratic examples while keeping the general graphical method the same.

2. The Linear-Linear Case: A Warm-Up

When both equations are linear, each graph is a straight line. Two straight lines either cross once (one solution), run parallel (no solution), or are the same line (infinitely many solutions). The most common exam case is a single crossing point.

The graph above shows $y = 2x + 1$ (a line with gradient 2 crossing the $y$-axis at 1) and $y = -x + 4$ (a line with gradient $-1$ crossing the $y$-axis at 4). They cross at one point — reading the coordinates from the graph gives the approximate solution to the simultaneous equations $y = 2x + 1$ and $y = -x + 4$.

Worked example: Reading a Linear-Linear Solution from a Graph

Use the graph of $y = 2x + 1$ and $y = -x + 4$ to find the approximate solution to the simultaneous equations.

1. The two lines cross at the point $(1, 3)$. — Trace along the grid carefully — the crossing is at $x = 1$ and $y = 3$. [M1 for identifying the intersection]
2. $x = 1, \quad y = 3$ — Both values must be stated. Writing only $x = 1$ loses the $y$ mark. [A1 for both values correct]
3. $y = 2(1) + 1 = 3$ ✓ $y = -(1) + 4 = 3$ ✓ — Substituting back confirms the solution is exact here. On a graph question, this check also catches misreads.

Answer: $x = 1$, $y = 3$

3. The Linear-Quadratic Case: Where It Gets Interesting

Now pair a straight line with a quadratic curve (a parabola). Because a parabola is curved, a straight line can cross it in three different ways: twice, once (as a tangent), or not at all. Each scenario gives a different number of solutions.

The higher-tier specification requires you to find approximate solutions using a graph for linear-quadratic systems. In practice, this means plotting both graphs carefully on the same axes and reading the $x$- and $y$-coordinates of each intersection point.

4. Plotting and Reading: The Full Method

The method for solving graphically has a clear sequence: build a table of values for each equation, plot both graphs on the same axes, then read off the intersection coordinates. Each stage earns method marks in an exam, so showing the table and the plotted points matters even if your final reading is slightly off.

Worked example: Solving a Linear-Quadratic System Graphically

Use a graphical method to find approximate solutions to the simultaneous equations $y = x^2 - 2x - 3$ and $y = x + 1$. Use values of $x$ from $-2$ to $4$.

1. For $y = x^2 - 2x - 3$: $x = -2$: $(-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5$ $x = -1$: $(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$ $x = 0$: $0 - 0 - 3 = -3$ $x = 1$: $1 - 2 - 3 = -4$ $x = 2$: $4 - 4 - 3 = -3$ $x = 3$: $9 - 6 - 3 = 0$ $x = 4$: $16 - 8 - 3 = 5$ — Expand each substitution fully — squaring first, then multiplying, then subtracting. [M1 for at least 4 correct $y$-values]
2. For $y = x + 1$: $x = -2$: $y = -1$ $x = 0$: $y = 1$ $x = 4$: $y = 5$ — A straight line only needs two points to plot, but three gives a useful check. [M1 for correct linear values]
3. Plot the seven points for the parabola and join with a smooth curve. Plot the three points for the line and join with a straight edge. — The parabola must be a smooth curve — not a series of straight line segments between points. The line must be drawn with a ruler. [M1 for correctly plotted graphs]
4. The line $y = x + 1$ crosses the parabola $y = x^2 - 2x - 3$ at two points. Reading from the graph: approximately $(-1, 0)$ and $(4, 5)$. — Read the $x$-coordinate first, then trace up or down to the $y$-coordinate. Both coordinates are needed for each solution. [A1 for each correct intersection point]
5. Solution 1: $x = -1, \quad y = 0$ Solution 2: $x = 4, \quad y = 5$ — These are the approximate solutions. You can verify: $(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$ ✓ and $(-1) + 1 = 0$ ✓. Both equations give $y = 0$ when $x = -1$, confirming the solution is exact in this case.

Answer: $x = -1, y = 0$ and $x = 4, y = 5$

Notice that the parabola is U-shaped, crossing the $x$-axis at $x = -1$ and $x = 3$ (its roots), with a minimum turning point at $x = 1$. The line cuts through it at two places, giving two distinct solutions. Reading those crossing points carefully from the grid is the core skill being assessed.

5. Reading Approximate Solutions Accurately

The word 'approximate' in the specification is deliberate. When intersection points fall between grid lines, you must estimate as carefully as possible. Examiners allow a tolerance — typically $\pm 0.1$ or $\pm 0.2$ — so a careful reading earns full marks even if it is not the exact algebraic answer.

Worked example: Reading an Approximate Solution Between Grid Lines

The graphs of $y = x^2 - x - 4$ and $y = 2x - 1$ are drawn on a grid. One intersection point appears to lie between $x = 2$ and $x = 3$, closer to $x = 3$. Estimate the solution at this intersection.

1. The crossing appears at approximately $x = 2.8$. — Estimate to one decimal place by judging how far between the grid lines the crossing sits. If it looks about four-fifths of the way from 2 to 3, write 2.8.
2. Substitute $x = 2.8$ into the linear equation (easier to compute): $y = 2(2.8) - 1 = 5.6 - 1 = 4.6$ — Using the linear equation avoids squaring a decimal, reducing arithmetic errors. You could also read $y$ directly from the graph.
3. $x \approx 2.8, \quad y \approx 4.6$ — The $\approx$ symbol signals this is an approximate reading. Examiners accept values within the stated tolerance of the exact answer.

Answer: $x \approx 2.8$, $y \approx 4.6$ (approximate, read from graph)

A useful shortcut when reading $y$: once you have estimated $x$ from the graph, substitute that value into whichever equation is easier to compute — usually the linear one. This gives a more reliable $y$-value than trying to read it directly off a curved graph.

6. Deliberate Practice

The worked examples above covered the method in full. Now it is your turn to apply it. These problems mirror the style of GCSE Higher exam questions — work through each one before checking the answer and solution.

7. Exam Technique and Common Mistakes

Graphical simultaneous equation questions follow a predictable mark scheme structure. Understanding where marks are awarded — and where they are commonly dropped — lets you approach these questions with a clear strategy.

One final point on exam language: if a question asks you to 'find approximate solutions using a graph', the mark scheme expects coordinates read from the graph, not an algebraic derivation. Showing algebra instead of graph-reading will not earn those marks, even if the algebra is correct.