One/no/infinite solutions | Year 10

01 / LEARN

The idea, step by step

Three possible relationships

Two distinct straight lines with different gradients cross once, giving one solution. Parallel distinct lines never meet, giving no solution. Equations describing the same line have infinitely many solutions: every point on that line satisfies both.

Read the algebraic conclusion

During elimination, a false statement such as $0 = 5$ proves inconsistency. A true identity such as $0 = 0$ means one equation supplied no new condition. For an identity, still describe the remaining line; it is not every point in the plane. Compare whole equations, not only their left sides.

A common trap

Interpreting $0 = 0$ as the single answer $x = 0, y = 0$ .

Instead: An identity does not fix either variable. Use the remaining equation to describe the common line.

02 / WORKED EXAMPLES

See the thinking

Example 1

No common point

Solve $2 x - y = 3$ and $4 x - 2 y = 8$ .

Doubling the first gives $4 x - 2 y = 6$ .
The second requires that same expression to equal $8$ .
Subtraction gives $0 = 2$ , a contradiction.

No solution

Example 2

The same line twice

Solve $x + y = 5$ and $2 x + 2 y = 10$ .

The second equation is twice the first.
Elimination gives $0 = 0$ , so there is no additional restriction.
Every point $(t, 5 - t)$ , for real $t$ , satisfies both.

Infinitely many: all points on

x + y = 5

Check your understanding

How many solutions do $y = 4 x + 1$ and $y = 4 x - 2$ have?

Show answer

None.

They have equal gradients but different intercepts, so they are parallel distinct lines.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Classify the solution set of $y = 2 x + 3$ and $y = - x + 6$ .
2 marks
Classify $x - y = 4$ and $2 x - 2 y = 8$ .
2 marks
Classify $3 x + y = 7$ and $6 x + 2 y = 9$ .
2 marks
Explain why $0 = 0$ does not mean that every pair of real numbers solves $x + y = 2$ and $2 x + 2 y = 4$ .
3 marks
Find the value of $k$ for which $y = 3 x + 5$ and $y = 3 x + k$ have infinitely many solutions.
2 marks
Two taxi companies charge the same 2 dollars per kilometre. Their starting fees are 4 and 7 dollars. Can the trip prices ever be equal for the same distance?
3 marks

Read the worked solutions

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

Classify the solution set of $y = 2 x + 3$ and $y = - x + 6$ .
2 marks
The gradients differ.
One solution.
Classify $x - y = 4$ and $2 x - 2 y = 8$ .
2 marks
The second equation is twice the first.
Infinitely many points on $x - y = 4$ .
Classify $3 x + y = 7$ and $6 x + 2 y = 9$ .
2 marks
Doubling the first would give right side $14$ , not $9$ .
No solution.
Explain why $0 = 0$ does not mean that every pair of real numbers solves $x + y = 2$ and $2 x + 2 y = 4$ .
3 marks
For example, $(0, 0)$ fails the original equation although the equations are dependent.
The original condition $x + y = 2$ still applies.
Find the value of $k$ for which $y = 3 x + 5$ and $y = 3 x + k$ have infinitely many solutions.
2 marks
Their intercepts must match.
$k = 5$
Two taxi companies charge the same 2 dollars per kilometre. Their starting fees are 4 and 7 dollars. Can the trip prices ever be equal for the same distance?
3 marks
The models are parallel lines separated by 3 dollars for every distance.
No.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian Mathematics Curriculum

Year 10 simultaneous equations and inequalities with graphical, algebraic and practical interpretations.

Official curriculum source ↗

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