Solving Linear Equations Graphically
An equation such as \(2x + 1 = -x + 7\) can be solved by graphing each side as a line and reading the \(x\)-coordinate where the lines cross. This NSW Year 9 Advanced maths subtopic solves linear equations with graphs (NESA outcome MA5‑LIN‑C‑01) and checks the answers algebraically.
You will draw pairs of lines, read solutions from graphs, and choose the horizontal line to draw for an equation like \(2x - 3 = 5\). You will explain why parallel lines mean no solution, estimate a crossing point and then find it exactly, compare two phone plans, find a cafe's break-even point, and find unknown constants from where two lines meet. It is part of Year 9 Advanced Coordinate Geometry and Graphs.
Every question with a fully worked solution.
- Solving Linear Equations Graphically - Video - Intersection Point Between a Horizontal and Vertical Line Watch
Theory
To solve an equation graphically, graph each side as a straight line. Where the lines cross, the two sides are equal, so the \(x\)-coordinate of the point of intersection is the solution. This Year 9 Advanced maths page covers horizontal lines for constants, reusing one graph, estimating then checking with algebra, parallel and identical lines, and break-even problems.
An equation such as \(5-2x=x-4\) says that two expressions are equal. Graph \(y=\) the left side and \(y=\) the right side on the same axes.
The lines meet at the point of intersection. There, both lines have the same \(y\)-value for the same \(x\), so the two sides are equal. The solution is the \(x\)-coordinate of that point, not the whole point.
A plain number on one side, such as \(=3\), is the horizontal line \(y=3\). A right side of \(0\) is the \(x\)-axis, so the solution is the \(x\)-intercept.
If the two lines are parallel (same gradient, different intercept), they never meet: the equation has no solution. If both sides give the same line, every \(x\) is a solution.
A graph can only be read as accurately as its grid. When the crossing is between grid lines, give an estimate and confirm it with algebra.
To solve \(f(x) = g(x)\), where both sides are linear:
| What the graphs do | Number of solutions |
|---|---|
| cross once (different gradients) | one |
| parallel (same \(m\), different \(c\)) | none |
| same line (same \(m\) and \(c\)) | infinitely many |
How to solve an equation graphically
- Split the equation: \(y=\) left side and \(y=\) right side.
- Graph both lines on the same axes, from tables of values or from \(m\) and \(c\).
- Read the point of intersection. The solution is its \(x\)-coordinate.
- Check by substituting into both sides of the original equation.
Variations
- Constant on one side: draw the horizontal line \(y=c\).
- Reusing a graph: rearrange so one side matches the line already drawn, then draw the horizontal line for the other side (Example 2).
- Off-grid crossing: estimate to one decimal place, then solve algebraically for the exact value (Example 3).
- Before you draw: compare gradients. Equal gradients mean no solution, or infinitely many if the intercepts also match.
- Unknown constant: if two lines cross at a given \(x\), substitute that \(x\) into both; equal \(y\)-values give an equation for the constant.
- Context: write each quantity as a line, find where they cross, and say which line is lower or higher on each side of the crossing (Example 4).
Graph each side of the equation as a line: \(y=5-2x\) and \(y=x-4\).
| \(x\) | \(0\) | \(2\) | \(4\) |
|---|---|---|---|
| \(5-2x\) | \(5\) | \(1\) | \(-3\) |
| \(x-4\) | \(-4\) | \(-2\) | \(0\) |
The lines cross at \((3,-1)\). Where they meet, the two sides are equal, so the solution is the \(x\)-coordinate: \(\boldsymbol{x=3}\).
| \(\text{Check: LHS}\) | \(=\) | \(5-2(3) = -1\) |
| \(\text{RHS}\) | \(=\) | \(3-4 = -1\ \checkmark\) |
(i) \(\dfrac{1}{2}x+1=3\)
(ii) \(\dfrac{x}{2}=-1\).
(i) The left side is the line already drawn. Draw \(y=3\) and read where they cross: \((4,3)\), so \(\boldsymbol{x=4}\).
(ii) Rewrite the equation so its left side matches the graph. Add \(1\) to both sides: \(\dfrac{1}{2}x+1=0\).
Now the right side is \(0\), and \(y=0\) is the \(x\)-axis. The line meets the \(x\)-axis at \((-2,0)\), so \(\boldsymbol{x=-2}\).
| \(\text{Check (ii): }\dfrac{-2}{2}\) | \(=\) | \(-1\ \checkmark\) |
(i) Estimate the solution of \(3x-1=-x+4\) from the graph.
(ii) Find the exact solution algebraically.
(i) The crossing is between \(x=1\) and \(x=2\), about a quarter of the way across: \(x\approx 1.2\) or \(1.3\).
(ii) Collect the \(x\)-terms on one side.
| \(3x+x-1\) | \(=\) | \(4\) |
| \(4x\) | \(=\) | \(5\) |
| \(x\) | \(=\) | \(\mathbf{\dfrac{5}{4}} = 1.25\) |
A graph gives a quick estimate; algebra gives the exact value. Both lines give \(y=2.75\) there, which confirms it.
(i) Write a cost equation for each gym.
(ii) For how many visits do they cost the same?
(iii) Which is cheaper for \(20\) visits?
(i) For \(n\) visits: \(C_A=15n\) and \(C_B=96+7n\).
(ii) Graph both lines; they cross where the costs are equal.
| \(15n\) | \(=\) | \(96+7n\) |
| \(8n\) | \(=\) | \(96\) |
| \(n\) | \(=\) | \(\mathbf{12}\) |
Both cost \(\$180\) for \(12\) visits.
(iii) After the crossing, B's line is lower. For \(20\) visits: \(C_A=\$300\) and \(C_B=96+140=\$236\). Gym B is cheaper.
Common pitfalls
Frequently asked questions
How do you solve an equation using a graph?
Graph the left-hand side and the right-hand side as two separate lines on the same axes. The \(x\)-coordinate of the point where they cross is the solution.
Why is the solution the x-coordinate of the intersection?
At the crossing point both lines give the same \(y\)-value for the same \(x\)-value. So for that \(x\), the left side equals the right side, which is what solving means.
What if the lines are parallel?
Parallel lines never cross, so the equation has no solution. This happens when both sides have the same gradient but different constants.
How accurate is solving an equation graphically?
Only as accurate as the grid and your drawing. If the crossing is between grid lines, give an estimate and confirm it with algebra.
What is a break-even point?
The point where two cost or income lines cross, so both options cost the same. On one side of it one option is cheaper; on the other side the other option is.