One/no/infinite solutions | Year 10

01 / LEARN

Understand the idea.

Three possible relationships

Two nonparallel straight lines intersect once, giving one solution. Distinct parallel lines have no common point, while coincident lines share every point on the line. Rewriting equations in gradient-intercept form can help identify which relationship occurs.

Read the algebraic conclusion

Elimination can identify these cases without drawing. If the variable terms cancel and leave a false statement such as $0 = 2$ , no pair can satisfy both equations. A contradiction is a conclusion, not an instruction to continue isolating a variable.

If elimination leaves $0 = 0$ , one equation repeats the other and supplies no extra restriction. Keep the remaining equation: infinitely many pairs lie on that line, but points elsewhere in the plane still fail the system.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

No common point

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

The lines y equals two x minus three and y equals two x minus four on equal-scale axes from zero to eight. They have the same slope and different intercepts, so they have no common point. Each line is identified by its equation.

Worked solution

1
Match the left sides.
$2 (2 x - y) = 2 (3) ⟹ 4 x - 2 y = 6$
Doubling the entire first equation preserves its solution line.
2
Compare it with the second equation.
$\begin{aligned} 4 x - 2 y & = 8 \\ 4 x - 2 y & = 6 \end{aligned}$
The same expression would have to equal two different constants.
3
Subtract to expose the contradiction.
$0 = 2$
No ordered pair can make this false equality true, so the system has no solution.

Answer

No solution

EXAMPLE 02

The same line twice

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

Worked solution

1
Double the first equation.
$2 (x + y) = 2 (5) ⟹ 2 x + 2 y = 10$
This is exactly the second equation, so both describe the same line.
2
Eliminate to identify the repeated condition.
$\begin{aligned} (2 x + 2 y) - (2 x + 2 y) & = 10 - 10 \\ 0 & = 0 \end{aligned}$
The identity adds no restriction beyond $x + y = 5$ .
3
Describe all pairs on the common line.
$(x, y) = (t, 5 - t), t \in ℝ$
Any real choice of $t$ gives $t + (5 - t) = 5$ , and doubling this equality satisfies the second equation.

Answer

Infinitely many: all points on $x + y = 5$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

None.

The two lines both have gradient $4$ but different intercepts. Equating their $y$ values gives $4 x + 1 = 4 x - 2$ , hence $1 = - 2$ , so they have no intersection and no solution.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Three possible relationships

Read the algebraic conclusion

See the method in action.

No common point

Match the left sides.

Compare it with the second equation.

Subtract to expose the contradiction.

The same line twice

Double the first equation.

Eliminate to identify the repeated condition.

Describe all pairs on the common line.

Try it yourself.

Check your understanding

Keep practising