Simultaneous linear equations and intersection models | Year 11

01 / LEARN

The idea, step by step

Satisfy both conditions

A simultaneous solution is a pair of values satisfying both equations at once. Graphically, it is an intersection. Substitution is useful when one variable is already isolated; elimination is useful when coefficients can be made equal or opposite. Neither method changes the requirement to check the final pair in both original equations.

When eliminating a variable, perform the whole operation on both sides. If you multiply an equation, multiply every term. After finding one variable, substitute it into a convenient original equation to obtain the other.

Interpret the number of solutions

Distinct parallel lines have no common point, so their equations have no solution. Two equations for the same line have infinitely many solutions. An algebraic contradiction such as 0=5 signals the first case; an identity such as 0=0 signals the second, not the single point (0,0).

In modelling, the algebraic solution may still be outside a meaningful domain. A negative number of tickets cannot solve a sales problem in practice. Report the quantities, units and restrictions along with the pair of numbers.

A common trap

Finding an intersection from a rough drawing and reporting it as an exact solution.

Instead: Use algebra for exact coordinates, then check both equations. A graph is valuable confirmation but its resolution can hide fractional intersections.

02 / WORKED EXAMPLES

See the thinking

Example 1

Substitution

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

Equate the expressions: $2 x - 1 = - x + 5$ , so 3x=6 and x=2.
Substitute to get y=3; both equations give that value.

(x, y) = (2, 3)

Example 2

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?

Let a+s=10 and 14a+9s=115.
Substitute s=10-a: 14a+90-9a=115, so a=5 and s=5.
Check: 5+5=10 and 70+45=115.

5 adult tickets and 5 student tickets.

Check your understanding

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

Show answer

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

Adding gives 2x=8; substitution gives y=3.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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 3x+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: 3x=12, then y=3.
$(4, 3)$
Solve $3 x + 2 y = 4$ and $x - y = 3$ .
2 marks
Use x=y+3: 3y+9+2y=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=2x-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+4v=6+7v 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 ↗

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