Not every system of simultaneous linear equations has a unique solution. Some have no solution (parallel lines), while others have infinite solutions (coincident lines). In DSE Paper 2 MC and Paper 1 Section A(1), you may be asked to identify the type of solution or find parameter values that lead to a particular case. This article teaches you the coefficient ratio method to quickly determine the solution type, with clear examples, common traps, and plenty of practice.
Learning Objectives
- Understand the three possible outcomes for a system of two linear equations.
- Use the coefficient ratio method to classify systems as unique, no, or infinite solutions.
- Determine the number of solutions from the given equations without solving.
- Find parameter values (e.g., \(k\)) that make a system have no solution or infinite solutions.
- Apply these skills to DSE-style multiple-choice and long questions.
Introduction
Consider a system of two linear equations:
\[
\begin{cases}
a_1 x + b_1 y = c_1 \\
a_2 x + b_2 y = c_2
\end{cases}
\]
When solving such a system, there are three possible outcomes:
- Unique solution: the lines intersect at one point.
- No solution: the lines are parallel and distinct.
- Infinite solutions: the lines are coincident (same line).
Instead of solving the system, you can determine the case by comparing the ratios of the coefficients:
\[
\frac{a_1}{a_2}, \quad \frac{b_1}{b_2}, \quad \frac{c_1}{c_2}.
\]
This method is especially useful in MC questions where you need a quick answer.
The Coefficient Ratio Method
For the system \(a_1 x + b_1 y = c_1\) and \(a_2 x + b_2 y = c_2\) (with \(a_2, b_2, c_2 \neq 0\)):
- Unique solution: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
- No solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
- Infinite solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
If any denominator is zero, you need to handle that separately (e.g., if \(a_2=0\), compare differently), but in DSE, denominators are usually non-zero.
Memory Trick
Think of it as: if the \(x\)-coefficient ratio equals the \(y\)-coefficient ratio, the lines are parallel (or coincident). If they are not equal, they intersect at one point. Then check the constant ratio to distinguish parallel (not equal) from coincident (equal).
Worked Examples (DSE-Style)
Example 1 – Unique Solution
Example 1
Determine the number of solutions for:
\[
\begin{cases}
2x + 3y = 5 \\
4x + 6y = 10
\end{cases}
\]
Solution:
Ratios: \(\frac{2}{4} = \frac{1}{2}\), \(\frac{3}{6} = \frac{1}{2}\), \(\frac{5}{10} = \frac{1}{2}\).
All ratios equal → infinite solutions (coincident lines).
Answer: Infinite solutions.
Example 2 – No Solution
Example 2
Determine the number of solutions for:
\[
\begin{cases}
3x - 2y = 7 \\
6x - 4y = 9
\end{cases}
\]
Solution:
Ratios: \(\frac{3}{6} = \frac{1}{2}\), \(\frac{-2}{-4} = \frac{1}{2}\), \(\frac{7}{9} \neq \frac{1}{2}\).
\(x\)-ratio = \(y\)-ratio ≠ constant ratio → no solution.
Answer: No solution.
Example 3 – Parameter for No Solution
Example 3
Find the value of \(k\) such that the system has no solution:
\[
\begin{cases}
x + 2y = 3 \\
2x + ky = 5
\end{cases}
\]
Solution:
For no solution: \(\frac{1}{2} = \frac{2}{k} \neq \frac{3}{5}\).
From \(\frac{1}{2} = \frac{2}{k}\) ⇒ \(k = 4\).
Check constant ratio: \(\frac{3}{5} \neq \frac{1}{2}\) (true).
Answer: \(k = 4\).
Example 4 – Parameter for Infinite Solutions
Example 4
Find \(k\) such that the system has infinite solutions:
\[
\begin{cases}
2x - 3y = 4 \\
kx - 6y = 8
\end{cases}
\]
Solution:
For infinite solutions: \(\frac{2}{k} = \frac{-3}{-6} = \frac{4}{8}\).
\(\frac{-3}{-6} = \frac{1}{2}\) and \(\frac{4}{8} = \frac{1}{2}\).
So \(\frac{2}{k} = \frac{1}{2}\) ⇒ \(k = 4\).
Answer: \(k = 4\).
DSE Exam Techniques – Paper 1 & Paper 2
Paper 1 Section A(1)
- If asked to determine the number of solutions, compare the ratios and state your conclusion clearly.
- For parameter problems, set up the appropriate ratio equations and solve.
- Show your working: write the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}\) and compare.
Paper 2 (MC)
- Quickly compute the ratios mentally to eliminate options.
- If the ratios are not equal, the answer is "unique solution" – no need to check constants.
- If the ratios are equal, check the constant ratio to decide between "no solution" and "infinite solutions".
DSE-Style Practice Questions
Determine the number of solutions for each system, or find the parameter as required.
Practice Set 1 – Identify the Type
- \(\begin{cases} 2x + 3y = 6 \\ 4x + 6y = 12 \end{cases}\)
- \(\begin{cases} 5x - 2y = 8 \\ 10x - 4y = 15 \end{cases}\)
- \(\begin{cases} 3x + y = 7 \\ 6x + 2y = 14 \end{cases}\)
Practice Set 2 – Parameter Problems
- Find \(k\) such that the system has no solution: \(\begin{cases} 2x + ky = 5 \\ 4x + 6y = 10 \end{cases}\)
- Find \(k\) such that the system has infinite solutions: \(\begin{cases} kx + 3y = 6 \\ 4x + 6y = 12 \end{cases}\)
- Find \(k\) such that the system has a unique solution: \(\begin{cases} 3x + 2y = 5 \\ kx + 4y = 8 \end{cases}\)
Practice Set 3 – Mixed
- \(\begin{cases} x - 2y = 3 \\ 2x - 4y = 6 \end{cases}\)
- \(\begin{cases} 4x + 3y = 7 \\ 8x + 6y = 13 \end{cases}\)
- Find \(k\) so that \(\begin{cases} 2x - y = 3 \\ 6x - 3y = k \end{cases}\) has no solution.
Solutions to Practice Sets
Set 1 Solutions
- \(\frac{2}{4} = \frac{3}{6} = \frac{6}{12} = \frac{1}{2}\) → Infinite solutions.
- \(\frac{5}{10} = \frac{-2}{-4} = \frac{1}{2}\), \(\frac{8}{15} \neq \frac{1}{2}\) → No solution.
- \(\frac{3}{6} = \frac{1}{2} = \frac{7}{14}\) → Infinite solutions.
Set 2 Solutions
- No solution: \(\frac{2}{4} = \frac{k}{6} \neq \frac{5}{10}\). \(\frac{2}{4} = \frac{1}{2}\), so \(\frac{k}{6} = \frac{1}{2} \Rightarrow k=3\). Check constant: \(\frac{5}{10} = \frac{1}{2}\) → equal, so infinite solutions, not no solution. Wait! If \(k=3\), then all ratios equal (1/2) → infinite. We need no solution: \(k\) such that \(x\)-ratio = \(y\)-ratio but constant different. So \(\frac{2}{4} = \frac{k}{6}\) gives \(k=3\), but then constant ratio is also 1/2, so infinite. So no solution is impossible for this system because the constants are proportional? Actually, if \(k=3\), it's infinite. For no solution, we need \(k\) such that \(x\)-ratio = \(y\)-ratio but constant ratio different. But \(x\)-ratio = 1/2, so \(k/6 = 1/2 \Rightarrow k=3\). Then constant ratio = 5/10 = 1/2, equal → infinite. So no value of \(k\) gives no solution. But maybe the question expects a value that makes it inconsistent? Let's re-examine: For no solution, \(\frac{2}{4} = \frac{k}{6}\) (so \(k=3\)) and \(\frac{5}{10} \neq \frac{2}{4}\). But \(\frac{5}{10} = \frac{1}{2}\) equals the ratio, so actually when \(k=3\), it's infinite. So there is no \(k\) for no solution. However, in typical DSE questions, they might have a different constant. I'll provide the correct reasoning: For no solution, we need \(\frac{2}{4} = \frac{k}{6}\) and \(\frac{5}{10} \neq \frac{2}{4}\). But \(\frac{5}{10} = \frac{1}{2}\) equals the ratio, so impossible. So answer: no such \(k\). I'll state that.
Correct solution: No solution requires \(\frac{2}{4} = \frac{k}{6} \Rightarrow k=3\), but then \(\frac{5}{10} = \frac{1}{2}\) equals, so infinite. Hence no value of \(k\) yields no solution.
- Infinite solutions: \(\frac{k}{4} = \frac{3}{6} = \frac{6}{12}\). \(\frac{3}{6} = \frac{1}{2}\), \(\frac{6}{12} = \frac{1}{2}\). So \(\frac{k}{4} = \frac{1}{2} \Rightarrow k=2\).
- Unique solution: \(\frac{3}{k} \neq \frac{2}{4}\). \(\frac{2}{4} = \frac{1}{2}\), so \(\frac{3}{k} \neq \frac{1}{2} \Rightarrow k \neq 6\).
Set 3 Solutions
- \(\frac{1}{2} = \frac{-2}{-4} = \frac{1}{2}\) and \(\frac{3}{6} = \frac{1}{2}\) → Infinite.
- \(\frac{4}{8} = \frac{3}{6} = \frac{1}{2}\), \(\frac{7}{13} \neq \frac{1}{2}\) → No solution.
- \(\begin{cases} 2x - y = 3 \\ 6x - 3y = k \end{cases}\). For no solution: \(\frac{2}{6} = \frac{-1}{-3} = \frac{1}{3}\), and constant ratio \(\frac{3}{k} \neq \frac{1}{3}\). So \(k \neq 9\). Also, if \(k=9\), infinite. So no solution for any \(k \neq 9\).
9. Exercise
Click the following link to have
An Exercise on "No Solution" vs "Infinite Solutions" in Simultaneous Equations
Summary Checklist for Revision
-
Unique solution: I can identify when \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
-
No solution: I know that \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
-
Infinite solutions: I recognise when all three ratios are equal.
-
Parameter problems: I can set up equations to find \(k\) for a given case.
-
Exam ready: I can apply this method quickly in Paper 2 MC.