Simultaneous linear equations and intersection models | Year 11

01 / LEARN

Understand the idea.

Satisfy both conditions

A simultaneous solution satisfies both equations at once. For two lines, it is their intersection. Use substitution when one variable is already isolated; use elimination when multiplying and combining equations can remove one variable.

Apply each algebraic operation to every term on both sides. After finding one variable, substitute into an original equation for the other. Checking the pair in both original equations tests the shared condition, rather than just one line.

Interpret the number of solutions

Distinct parallel lines have no solution; equations describing the same line have infinitely many. A contradiction such as $0 = 5$ signals no common point. An identity such as $0 = 0$ leaves a whole line of possibilities rather than fixing the origin.

A contextual solution must also satisfy the quantities’ restrictions. Ticket counts must be non-negative integers, for example. Define the variables, set up both constraints, and report the answer with its meaning and units.

The sage line y=2x-1 and gold line y=-x+5 are shown in the original x=-1 to 5 and y=-1 to 7 window. Their intersection is not newly labelled.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Substitution

Solve $y = 2 x - 1$ and $y = - x + 5$ .

Worked solution

1
Equate the two expressions for the same output.
$2 x - 1 = - x + 5$
At an intersection, both rules must give the same y-value.
2
Solve for the input.
$\begin{aligned} 3 x - 1 & = 5 \\ 3 x & = 6 \\ x & = 2 \end{aligned}$
Adding x, then one, and dividing by three preserves equality.
3
Find and check the shared output.
$\begin{aligned} y & = 2 (2) - 1 = 3 \\ - 2 + 5 & = 3 \end{aligned}$
Both original rules give three at input two, so the intersection is the point with coordinates two and three.

Answer

$(x, y) = (2, 3)$

EXAMPLE 02

A ticket model

Adult tickets cost 14 dollars and student tickets cost 9 dollars. Ten tickets bring in 115 dollars. How many of each were sold?

Worked solution

1
Translate the count and revenue constraints.
$\begin{aligned} a + s & = 10 \\ 14 a + 9 s & = 115 \end{aligned}$
Let a count adult tickets and s count student tickets; revenue is price multiplied by count.
2
Eliminate the student count by substitution.
$\begin{aligned} s & = 10 - a \\ 14 a + 9 (10 - a) & = 115 \\ 5 a + 90 & = 115 \end{aligned}$
The total-ticket equation expresses the student count in terms of the adult count.
3
Solve for both counts.
$\begin{aligned} 5 a & = 25 \\ a & = 5 \\ s & = 10 - 5 = 5 \end{aligned}$
Solving the reduced equation and substituting back gives whole, non-negative counts.
4
Verify both constraints.
$5 + 5 = 10, 14 (5) + 9 (5) = 115$
Five tickets of each type match both the total number and the total revenue.

Answer

5 adult tickets and 5 student tickets.

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $x + y = 7$ and $x - y = 1$ .

Show answer & reasoning

$x = 4$ , $y = 3$ .

Adding the equations gives $2 x = 8$ , hence $x = 4$ . Then $y = 7 - 4 = 3$ ; the check $4 - 3 = 1$ confirms the second equation as well.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Solve $y = 3 x + 1$ and $y = x + 9$ .
2 marks
Solve $2 x + y = 11$ and $x - y = 1$ .
2 marks
Solve $3 x + 2 y = 4$ and $x - y = 3$ .
2 marks
Classify $y = 2 x + 1$ and $4 x - 2 y = 6$ by their number of simultaneous solutions.
2 marks
Classify $x + 2 y = 5$ and $3 x + 6 y = 15$ .
2 marks
Plan A costs 18 dollars plus 4 dollars per visit; Plan B costs 6 dollars plus 7 dollars per visit. At what number of visits are their costs equal?
2 marks

Read the worked solutions

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

Solve $y = 3 x + 1$ and $y = x + 9$ .
2 marks
Set $3 x + 1 = x + 9$ to obtain $x = 4$ , then $y = 13$ .
$(4, 13)$
Solve $2 x + y = 11$ and $x - y = 1$ .
2 marks
Add equations: $3 x = 12$ , then $y = 3$ .
$(4, 3)$
Solve $3 x + 2 y = 4$ and $x - y = 3$ .
2 marks
Use $x = y + 3$ : $3 y + 9 + 2 y = 4$ , so $y = - 1$ and $x = 2$ .
$(2, - 1)$
Classify $y = 2 x + 1$ and $4 x - 2 y = 6$ by their number of simultaneous solutions.
2 marks
The second is $y = 2 x - 3$ : distinct parallel lines.
No solutions.
Classify $x + 2 y = 5$ and $3 x + 6 y = 15$ .
2 marks
The second is three times the first; both describe one line.
Infinitely many solutions.
Plan A costs 18 dollars plus 4 dollars per visit; Plan B costs 6 dollars plus 7 dollars per visit. At what number of visits are their costs equal?
2 marks
Solve $18 + 4 v = 6 + 7 v$ to obtain $v = 4$ .
Check both original costs.
4 visits; both cost 34 dollars.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Topic 1: Functions and graphs. Prerequisite algebra and coordinate geometry are embedded as supporting lessons.

Official curriculum source ↗

In this lesson

Understand the idea.

Satisfy both conditions

Interpret the number of solutions

See the method in action.

Substitution

Equate the two expressions for the same output.

Solve for the input.

Find and check the shared output.

A ticket model

Translate the count and revenue constraints.

Eliminate the student count by substitution.

Solve for both counts.

Verify both constraints.

Try it yourself.

Check your understanding

Keep practising