Graphical solutions and intersection checks | Year 8

01 / LEARN

Understand the idea.

One point satisfies both rules

An intersection is an ordered pair that satisfies both graph rules. Its first coordinate is a common input and its second is their common output. Give both coordinates when identifying the point.

Plotting can suggest where two lines meet, but a grid reading may only be approximate. At an exact intersection, the two output expressions are equal; solve that equality and substitute back to find the output.

Recognise when a picture is approximate

A crossing between grid lines may have fractional coordinates. Report an approximate graph reading honestly, then use algebra if you need an exact value. Checking both original rules verifies that the proposed point lies on each.

Two different parallel lines never meet because they have equal rates and different vertical-axis crossings. Identical lines overlap everywhere, so every point on the line is common rather than there being one intersection.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A visible common point

Find the intersection of $y = 2 x + 1$ and $y = - x + 4$ .

Equal-scale Cartesian axes from negative 4 to 4 show the sage line y equals 2x plus 1 and the gold line y equals negative x plus 4, each clipped at the grid boundary. No intersection coordinates are labelled.

Worked solution

1
Set the output expressions equal.
$2 x + 1 = - x + 4$
At an intersection, both rules must assign the same output to the same input.
2
Gather terms and solve for the input.
$\begin{aligned} 3 x + 1 & = 4 \\ 3 x & = 3 \\ x & = 1 \end{aligned}$
Add $x$ to both sides, subtract $1$ , then divide by $3$ .
3
Find and check the common output.
$\begin{aligned} 2 (1) + 1 & = 3 \\ - (1) + 4 & = 3 \end{aligned}$
Both original rules produce the same height.
4
State the intersection as a point.
$(1, 3)$
The input is the first coordinate and the common output is the second.

Answer

$(1, 3)$ .

EXAMPLE 02

A horizontal partner

Find the intersection of $y = x + 2$ and $y = 4$ .

Worked solution

1
Use the horizontal line's fixed output.
$y = 4$
Every point on the second graph has height four.
2
Put that output into the other rule.
$4 = x + 2$
The common point must also satisfy the first equation.
3
Subtract 2 and verify the output.
$\begin{aligned} x & = 2 \\ 2 + 2 & = 4 \end{aligned}$
This input gives the required height in the first rule as well.
4
Write the ordered pair.
$(2, 4)$
Both original equations hold at this point.

Answer

$(2, 4)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Does $(2, 5)$ lie on both $y = 2 x + 1$ and $y = x + 3$ ?

Show answer & reasoning

Yes.

At $x = 2$ , the first rule gives $2 (2) + 1 = 5$ , and the second gives $2 + 3 = 5$ . Both match the proposed second coordinate, so $(2, 5)$ lies on both graphs.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Verify that $(1, 4)$ lies on $y = 3 x + 1$ and $y = x + 3$ .
2 marks
Find the intersection of $y = x + 1$ and $y = 3$ .
2 marks
Find the intersection of $y = 2 x$ and $y = x + 2$ .
2 marks
Find the intersection of $y = - x + 5$ and $y = x + 1$ .
2 marks
A crossing of $y = 2 x + 1$ and $y = x + 3$ is read as $(2, 6)$ . Use checks to assess this.
3 marks
Can $y = 2 x + 1$ and $y = 2 x - 3$ intersect? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Verify that $(1, 4)$ lies on $y = 3 x + 1$ and $y = x + 3$ .
2 marks
First output: $3 + 1 = 4$ .
Second output: $1 + 3 = 4$ .
It lies on both.
Find the intersection of $y = x + 1$ and $y = 3$ .
2 marks
Solve $x + 1 = 3$ .
Then $x = 2$ and $y = 3$ .
$(2, 3)$ .
Find the intersection of $y = 2 x$ and $y = x + 2$ .
2 marks
Set $2 x = x + 2$ , giving $x = 2$ .
The shared output is 4.
$(2, 4)$ .
Find the intersection of $y = - x + 5$ and $y = x + 1$ .
2 marks
Set $- x + 5 = x + 1$ , giving $2 x = 4$ .
Substitute $x = 2$ to get $y = 3$ .
$(2, 3)$ .
A crossing of $y = 2 x + 1$ and $y = x + 3$ is read as $(2, 6)$ . Use checks to assess this.
3 marks
Both rules give 5 at $x = 2$ .
The proposed $x$ works, but the $y$ -reading is one unit too high.
The correct point is $(2, 5)$ .
Can $y = 2 x + 1$ and $y = 2 x - 3$ intersect? Explain.
3 marks
At every input the first output is 4 larger.
The graphs are parallel distinct lines.
No.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics (2026)

Multi-step equations; brackets and variables on both sides; equations with rational solutions; substitution checks; one-variable inequalities and number-line solutions; tables and straight-line graphs; horizontal/vertical lines; how coefficients change a graph; graphical solutions and intersection checks; real-world linear models with digital exploration.

Official curriculum source ↗

In this lesson

Understand the idea.

One point satisfies both rules

Recognise when a picture is approximate

See the method in action.

A visible common point

Set the output expressions equal.

Gather terms and solve for the input.

Find and check the common output.

State the intersection as a point.

A horizontal partner

Use the horizontal line's fixed output.

Put that output into the other rule.

Subtract 2 and verify the output.

Write the ordered pair.

Try it yourself.

Check your understanding

Keep practising