Handling Simultaneous Equations with Fractions or Decimals

Published: 2026-09-04 Categories: Equations and Identities Editor: MathSir Difficulty: ★★★★☆ Reading time: ~14 minutes Pageview: 41

Synopsis

Simultaneous equations often contain fractions or decimals, which can make them look intimidating. However, the key is to clear the fractions or decimals first by multiplying each equation by the LCM of denominators or by a power of 10. This transforms the system into one with integer coefficients, which you can then solve using substitution or elimination. This article provides a step-by-step strategy for handling these equations, with worked examples, common pitfalls, and plenty of DSE-style practice.


Target Exam Paper Paper 1 Section A(1) & A(2)
Typical Marks 4–6 marks
Syllabus Ref. Number and Algebra – Simultaneous linear equations, algebraic fractions

Learning Objectives

  • Identify simultaneous equations with fractions or decimals.
  • Clear fractions by multiplying each equation by the least common multiple (LCM) of the denominators.
  • Clear decimals by multiplying each equation by a power of 10 (e.g., 10, 100).
  • Solve the resulting integer-coefficient systems using substitution or elimination.
  • Check solutions by substituting back into the original equations.

Introduction

In DSE Paper 1, you may encounter simultaneous equations like: \[ \begin{cases} \frac{x}{2} + \frac{y}{3} = 5 \\ \frac{x}{4} - \frac{y}{2} = 1 \end{cases} \quad \text{or} \quad \begin{cases} 0.2x + 0.3y = 1.4 \\ 0.5x - 0.1y = 0.7 \end{cases} \] These can be solved by the usual substitution or elimination methods, but the fractions or decimals make calculations messy. The solution is to transform the system into one with integer coefficients by multiplying each equation by a suitable number.

This article will teach you the techniques for clearing fractions and decimals, with step-by-step examples and tips to avoid common errors.

Technique 1: Clearing Fractions

To clear fractions, multiply every term of both equations by the LCM of the denominators in that equation. (Note: you can use a different multiplier for each equation if the denominators are different.)

Key Point

Multiply both sides of the equation by the LCM. For example, if the denominators are 2 and 3, the LCM is 6. Multiply every term by 6.

Worked Example 1 – Clearing Fractions

Example 1

Solve the system: \[ \begin{cases} \frac{x}{2} + \frac{y}{3} = 5 \\ \frac{x}{4} - \frac{y}{2} = 1 \end{cases} \]

Solution:

Equation 1: denominators 2 and 3 → LCM = 6. Multiply by 6:
\(6 \cdot \frac{x}{2} + 6 \cdot \frac{y}{3} = 6 \cdot 5 \Rightarrow 3x + 2y = 30\).
Equation 2: denominators 4 and 2 → LCM = 4. Multiply by 4:
\(4 \cdot \frac{x}{4} - 4 \cdot \frac{y}{2} = 4 \cdot 1 \Rightarrow x - 2y = 4\).
Now solve the system: \[ \begin{cases} 3x + 2y = 30 \\ x - 2y = 4 \end{cases} \] Add the equations: \(4x = 34 \Rightarrow x = \frac{17}{2} = 8.5\).
Substitute into \(x - 2y = 4\): \(8.5 - 2y = 4 \Rightarrow -2y = -4.5 \Rightarrow y = 2.25 = \frac{9}{4}\).
Check in original equations: works.
Answer: \(x = \frac{17}{2}, y = \frac{9}{4}\).

Worked Example 2 – Fractions with Variables in Denominators

Example 2

Solve the system: \[ \begin{cases} \frac{2}{x} + \frac{3}{y} = 7 \\ \frac{1}{x} - \frac{1}{y} = 1 \end{cases} \]

Solution:

This is a system where the variables appear in the denominators. Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\).
Then the system becomes: \[ \begin{cases} 2u + 3v = 7 \\ u - v = 1 \end{cases} \] From the second equation, \(u = v + 1\).
Substitute: \(2(v+1) + 3v = 7 \Rightarrow 2v+2+3v=7 \Rightarrow 5v=5 \Rightarrow v=1\).
Then \(u = 1+1=2\).
So \(\frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}\), and \(\frac{1}{y} = 1 \Rightarrow y = 1\).
Answer: \(x = \frac{1}{2}, y = 1\).

Technique 2: Clearing Decimals

To clear decimals, multiply both sides of each equation by a power of 10 (10, 100, 1000, etc.) so that all numbers become integers. Choose the smallest power that eliminates all decimals in that equation.

Key Point

Count the maximum number of decimal places in the equation. If the longest decimal has 2 places, multiply by 100. If it has 1 place, multiply by 10.

Worked Example 3 – Clearing Decimals

Example 3

Solve the system: \[ \begin{cases} 0.2x + 0.3y = 1.4 \\ 0.5x - 0.1y = 0.7 \end{cases} \]

Solution:

Equation 1: decimals to 1 place → multiply by 10: \(2x + 3y = 14\).
Equation 2: decimals to 1 place → multiply by 10: \(5x - y = 7\).
Now solve: \[ \begin{cases} 2x + 3y = 14 \\ 5x - y = 7 \end{cases} \] From the second equation, \(y = 5x - 7\).
Substitute: \(2x + 3(5x - 7) = 14 \Rightarrow 2x + 15x - 21 = 14 \Rightarrow 17x = 35 \Rightarrow x = \frac{35}{17}\).
Then \(y = 5(\frac{35}{17}) - 7 = \frac{175}{17} - \frac{119}{17} = \frac{56}{17}\).
Answer: \(x = \frac{35}{17}, y = \frac{56}{17}\).

General Strategy & DSE Exam Techniques

Step-by-Step Approach
  1. If the equation contains fractions, find the LCM of the denominators and multiply every term by it.
  2. If the equation contains decimals, multiply by 10, 100, etc., to make all numbers integers.
  3. If variables appear in denominators, use substitution (e.g., let \(u = 1/x\), \(v = 1/y\)).
  4. Solve the resulting integer-coefficient system using substitution or elimination.
  5. Check your solution in the original equations to avoid errors.
Paper 1 Section A(1) & A(2)
  • Show the step where you multiply to clear fractions or decimals – this gains method marks.
  • Be careful: when multiplying, multiply every term, including constants on the right-hand side.
  • For fractions with variables in denominators, remember to state any restrictions (e.g., \(x \neq 0, y \neq 0\)) if asked.

DSE-Style Practice Questions

Solve each system. Clear fractions or decimals first.

Practice Set 1 – Fractions
  1. \(\begin{cases} \frac{x}{3} + \frac{y}{2} = 4 \\ \frac{x}{6} - \frac{y}{4} = 1 \end{cases}\)
  2. \(\begin{cases} \frac{2x}{5} - \frac{y}{3} = 2 \\ \frac{x}{10} + \frac{y}{6} = 1 \end{cases}\)
  3. \(\begin{cases} \frac{x+1}{2} + \frac{y-1}{3} = 2 \\ \frac{x-1}{3} - \frac{y+1}{2} = 0 \end{cases}\)
Practice Set 2 – Decimals
  1. \(\begin{cases} 0.3x + 0.2y = 1.6 \\ 0.5x - 0.1y = 0.7 \end{cases}\)
  2. \(\begin{cases} 0.4x - 0.3y = 0.1 \\ 0.2x + 0.5y = 1.1 \end{cases}\)
  3. \(\begin{cases} 0.05x + 0.03y = 0.29 \\ 0.02x - 0.07y = -0.03 \end{cases}\)
Practice Set 3 – Variables in Denominators
  1. \(\begin{cases} \frac{3}{x} + \frac{2}{y} = 8 \\ \frac{1}{x} - \frac{1}{y} = 1 \end{cases}\)
  2. \(\begin{cases} \frac{2}{x} + \frac{5}{y} = 9 \\ \frac{3}{x} - \frac{2}{y} = 4 \end{cases}\)
  3. \(\begin{cases} \frac{4}{x} - \frac{3}{y} = 2 \\ \frac{2}{x} + \frac{9}{y} = 5 \end{cases}\)

Solutions to Practice Sets

Set 1 Solutions
  1. LCMs: 6 and 12. Multiply first by 6: \(2x+3y=24\). Second by 12: \(2x-3y=12\). Add: \(4x=36 \Rightarrow x=9\). Then \(2(9)+3y=24 \Rightarrow 18+3y=24 \Rightarrow y=2\).
  2. LCMs: 15 and 30. First by 15: \(6x-5y=30\). Second by 30: \(3x+5y=30\). Add: \(9x=60 \Rightarrow x=\frac{20}{3}\). Then \(6(\frac{20}{3})-5y=30 \Rightarrow 40-5y=30 \Rightarrow y=2\).
  3. First by 6: \(3(x+1)+2(y-1)=12 \Rightarrow 3x+3+2y-2=12 \Rightarrow 3x+2y=11\). Second by 6: \(2(x-1)-3(y+1)=0 \Rightarrow 2x-2-3y-3=0 \Rightarrow 2x-3y=5\). Solve: multiply first by 3: \(9x+6y=33\), second by 2: \(4x-6y=10\). Add: \(13x=43 \Rightarrow x=\frac{43}{13}\). Then \(2(\frac{43}{13})-3y=5 \Rightarrow \frac{86}{13}-3y=5 \Rightarrow -3y = \frac{65}{13}-\frac{86}{13} = -\frac{21}{13} \Rightarrow y=\frac{7}{13}\).
Set 2 Solutions
  1. Multiply by 10: \(3x+2y=16\), \(5x-y=7\). From second \(y=5x-7\). Substitute: \(3x+2(5x-7)=16 \Rightarrow 3x+10x-14=16 \Rightarrow 13x=30 \Rightarrow x=\frac{30}{13}\). Then \(y=5(\frac{30}{13})-7 = \frac{150}{13}-\frac{91}{13}=\frac{59}{13}\).
  2. Multiply by 10: \(4x-3y=1\), \(2x+5y=11\). Multiply second by 2: \(4x+10y=22\). Subtract first: \((4x+10y)-(4x-3y)=22-1 \Rightarrow 13y=21 \Rightarrow y=\frac{21}{13}\). Then \(4x-3(\frac{21}{13})=1 \Rightarrow 4x = 1+\frac{63}{13} = \frac{76}{13} \Rightarrow x=\frac{19}{13}\).
  3. Multiply by 100: \(5x+3y=29\), \(2x-7y=-3\). Solve: multiply first by 7: \(35x+21y=203\), second by 3: \(6x-21y=-9\). Add: \(41x=194 \Rightarrow x=\frac{194}{41}\). Then \(5(\frac{194}{41})+3y=29 \Rightarrow \frac{970}{41}+3y = \frac{1189}{41} \Rightarrow 3y = \frac{219}{41} \Rightarrow y=\frac{73}{41}\).
Set 3 Solutions
  1. Let \(u=1/x, v=1/y\). Then \(3u+2v=8\), \(u-v=1\). From second \(u=v+1\). Substitute: \(3(v+1)+2v=8 \Rightarrow 5v+3=8 \Rightarrow v=1\). Then \(u=2\). So \(x=1/2, y=1\).
  2. Let \(u=1/x, v=1/y\). \(2u+5v=9\), \(3u-2v=4\). Multiply first by 2: \(4u+10v=18\), second by 5: \(15u-10v=20\). Add: \(19u=38 \Rightarrow u=2\). Then \(2(2)+5v=9 \Rightarrow 4+5v=9 \Rightarrow v=1\). So \(x=1/2, y=1\).
  3. Let \(u=1/x, v=1/y\). \(4u-3v=2\), \(2u+9v=5\). Multiply first by 3: \(12u-9v=6\), add to second: \(14u=11 \Rightarrow u=\frac{11}{14}\). Then \(2(\frac{11}{14})+9v=5 \Rightarrow \frac{11}{7}+9v=5 \Rightarrow 9v = \frac{35}{7}-\frac{11}{7}=\frac{24}{7} \Rightarrow v=\frac{8}{21}\). So \(x=\frac{14}{11}, y=\frac{21}{8}\).

9. Exercise

Click the following link to have
 An Exercise on Handling Simultaneous Equations with Fractions or Decimals

Summary Checklist for Revision

  • Fractions: I can find the LCM of denominators and multiply every term to clear them.
  • Decimals: I can multiply by a power of 10 to make all numbers integers.
  • Variables in denominators: I can substitute \(u=1/x, v=1/y\) to simplify.
  • Solve: I can use substitution or elimination on the transformed system.
  • Check: I always substitute back into the original equations.

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