Solving simultaneous equations graphically/algebraically/with technology | Year 12

01 / LEARN

The idea, step by step

Satisfy both conditions

A simultaneous solution works in both equations at once. On a graph it is a shared point. In substitution, replace one variable using an equivalent expression from the other equation. In elimination, add suitable multiples of the equations so that one variable disappears, then recover the other value.

Check a graph or calculator result

A graphing tool can locate an intersection, but window settings and rounding can hide its exact coordinates. Substitute the pair into both original equations. Distinct parallel lines have no solution. Equivalent equations represent the same line and therefore have infinitely many solutions; a calculator error message alone is not the explanation.

A common trap

One equation is satisfied, so the pair is a solution.

Instead: The pair must satisfy every equation, not just the first.

02 / WORKED EXAMPLES

See the thinking

Example 1

Elimination

Solve $2 x + y = 11$ and $x - y = 1$ .

Add the equations to obtain $3 x = 12$ , so $x = 4$ .
Substitute into the second: $4 - y = 1$ , so $y = 3$ .
Check $8 + 3 = 11$ and $4 - 3 = 1$ .

(4, 3)

Example 2

An intersection

Find where $y = 2 x + 1$ meets $y = - x + 7$ .

Set $2 x + 1 = - x + 7$ .
Then $3 x = 6$ , $x = 2$ and $y = 5$ .
Both original rules give 5 at $x = 2$ .

(2, 5)

Check your understanding

Solve $x + y = 9$ and $x - y = 3$ .

Show answer

$(6, 3)$ .

Add to get $2 x = 12$ , then substitute.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Solve $x + y = 13$ and $x - y = 5$ .
2 marks
Solve $3 x + 2 y = 16$ and $x - y = 2$ .
2 marks
Find where $y = 4 x - 3$ meets $y = x + 6$ .
2 marks
How many solutions do $x + 2 y = 6$ and $2 x + 4 y = 15$ have?
2 marks
How many solutions do $3 x - y = 8$ and $6 x - 2 y = 16$ have?
2 marks
A solver returns $(2, 5)$ for $4 x - y = 3$ and $x + 3 y = 17$ . Verify it.
2 marks

Read the worked solutions

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

Solve $x + y = 13$ and $x - y = 5$ .
2 marks
Adding gives $2 x = 18$ ; then $y = 4$ .
$(9, 4)$ .
Solve $3 x + 2 y = 16$ and $x - y = 2$ .
2 marks
Substitute $x = y + 2$ : $5 y + 6 = 16$ .
Thus $y = 2$ , $x = 4$ .
$(4, 2)$ .
Find where $y = 4 x - 3$ meets $y = x + 6$ .
2 marks
$4 x - 3 = x + 6$ gives $x = 3$ ; then $y = 9$ .
$(3, 9)$ .
How many solutions do $x + 2 y = 6$ and $2 x + 4 y = 15$ have?
2 marks
Doubling the first requires $2 x + 4 y = 12$ , which contradicts 15.
None.
How many solutions do $3 x - y = 8$ and $6 x - 2 y = 16$ have?
2 marks
The second equation is twice the first, so they describe the same line.
Infinitely many.
A solver returns $(2, 5)$ for $4 x - y = 3$ and $x + 3 y = 17$ . Verify it.
2 marks
$4 (2) - 5 = 3$ and $2 + 3 (5) = 17$ .
Correct.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: Modelling with Linear Relationships, a compulsory school-assessed topic. The external examination scope differs from the full course.

Official curriculum source ↗

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