A fractional equation often looks more complicated than the equation hidden inside it. The key move is to identify denominator restrictions, clear the denominators correctly, solve the resulting linear or quadratic equation, then return to the original equation to reject any inadmissible roots.
This thirty-fifth Secondary 4 Mathematics Learning Guide develops fractional equations as a bridge between algebraic fractions and ordinary equation solving. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
It builds on Algebraic Fractions: Multiply, Divide, Add and Subtract and Quadratic Equations and Algebraic Fractions.
Step 1: identify values that make denominators zero
Before multiplying through by denominators, record excluded values. For:
3/(x−2)=5
the original restriction is x≠2.
This restriction remains even if later algebra produces x=2 as a formal candidate.
Worked Example 1 | Fractional equation reduced to linear form
Solve 3/(x−2)=5.
Restriction: x≠2.
Multiply both sides by x−2:
3=5(x−2).
3=5x−10, so 5x=13.
x=13/5.
13/5 is allowed and satisfies the original equation.
Clearing denominators is multiplication, not cancellation across addition
For an equation such as 2/x + 1/(x+1)=3, multiply every term by the common denominator x(x+1). Do not cancel pieces independently across a sum.
Worked Example 2 | Two fractions reduce to a quadratic
Solve 2/x + 1/(x+1)=3.
Restrictions: x≠0,−1.
Multiply through by x(x+1):
2(x+1)+x=3x(x+1).
3x+2=3x²+3x.
3x²−2=0.
x=±√(2/3).
Neither root is excluded by the original denominators, so both are admissible real solutions.
Worked Example 3 | Linear reduction after common denominator
Solve 4/(x−1)−1/(x−1)=6.
Restriction: x≠1.
The fractions already share a denominator:
3/(x−1)=6.
Multiply by x−1:
3=6(x−1), so x−1=1/2.
x=3/2.
A candidate root can be invalid
When denominators contain x, multiplication by an expression involving x can transform the equation into a polynomial that has roots not permitted in the original domain.
This is why restrictions must be written before clearing denominators.
Worked Example 4 | Reject an excluded root
Solve x/(x−2)=2/(x−2).
Restriction: x≠2.
Multiply by x−2:
x=2.
But x=2 is excluded by the original denominator.
No solution.
Factor denominators before choosing a common multiple
If one denominator is x²−4, rewrite it as (x−2)(x+2). This may reveal that another denominator such as x−2 is already contained in it.
Worked Example 5 | Factored denominator
Solve 1/(x²−4)+1/(x−2)=1.
Restrictions: x≠2,−2.
Factor x²−4=(x−2)(x+2). Multiply through by (x−2)(x+2):
1+(x+2)=(x−2)(x+2).
x+3=x²−4.
x²−x−7=0.
Using the quadratic formula:
x=(1±√29)/2.
Neither value equals ±2, so both are admissible.
Worked Example 6 | Equation with a constant plus a fraction
Solve 2+3/x=5.
Restriction: x≠0.
Subtract 2:
3/x=3.
x=1.
Sometimes isolating the fraction before clearing denominators is the shortest route.
Cross multiplication is a special case
If A/B=C/D and both denominators are non-zero, then AD=BC. This is simply the result of multiplying both sides by BD.
Cross multiplication is therefore safe only when the equation is genuinely one fraction equal to one fraction.
Worked Example 7 | One fraction equals another
Solve (x+1)/(x−2)=3/2.
Restriction: x≠2.
2(x+1)=3(x−2).
2x+2=3x−6.
x=8.
Check: 9/6=3/2.
Fractional equations can model rates
Time often appears as distance divided by speed. Comparing journey times can therefore create fractional equations.
Worked Example 8 | Travel-time equation
A 120 km journey at speed v km/h takes one hour longer than the same journey at v+20 km/h. Find v.
Assume positive speeds and v≠0,−20.
120/v − 120/(v+20)=1.
Multiply through by v(v+20):
120(v+20)−120v=v(v+20).
2400=v²+20v.
v²+20v−2400=0.
(v+60)(v−40)=0.
The physical context requires positive speed, so:
v=40 km/h.
Check: 120/40=3 h and 120/60=2 h.
Fractional equations can model work rates
If one machine completes a job in a hours, its idealised work rate is 1/a job per hour. Combined rates add when the model assumes independent constant work rates.
Worked Example 9 | Combined work-rate model
Machine A completes a job in x hours. Machine B completes the same job in x+3 hours. Together they complete one job in 2 hours. Form the equation.
1/x + 1/(x+3)=1/2.
Restrictions and context require x>0.
Multiply through by 2x(x+3):
2(x+3)+2x=x(x+3).
4x+6=x²+3x.
x²−x−6=0=(x−3)(x+2).
Context rejects x=−2, leaving x=3 hours.
Verification should return to the original equation
Checking only the transformed polynomial is not enough. A root can solve the transformed equation yet fail the original denominator restriction.
A reliable check is:
- Confirm the root is not excluded.
- Substitute into the original fractional equation.
- Confirm both sides are defined and equal.
Worked Example 10 | Full verification
Solve 1/(x−1)=x−1.
Restriction: x≠1.
Multiply by x−1:
1=(x−1)².
x−1=±1, so x=2 or x=0.
Both are permitted.
Check x=2: 1/(1)=1 and x−1=1.
Check x=0: 1/(−1)=−1 and x−1=−1.
x=0 or x=2.
A disciplined workflow
- Factor denominators where useful.
- Record all excluded values.
- Choose the lowest useful common denominator.
- Multiply every term by it.
- Simplify to a linear or quadratic equation.
- Solve accurately.
- Reject excluded or context-invalid roots.
- Substitute surviving roots into the original equation.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Clears denominator from only one term | Common multiplier not applied to whole equation | Multiply every term |
| Cross-multiplies a sum of fractions | Special-case rule overused | Use an LCD for multi-term equations |
| Accepts denominator-zero root | Restrictions recorded too late | Write exclusions before solving |
| Stops at transformed quadratic | Original equation forgotten | Verify candidates in the original form |
| Keeps negative speed or time | Context not reapplied | Filter roots through physical meaning |
| Expands denominators before factoring | Structure hidden | Factor first to find a smaller LCD |
Independent practice
- Solve 5/(x+1)=2.
- Solve 1/x+1/(x+2)=1.
- Solve x/(x−4)=4/(x−4).
- Solve (x+2)/(x−1)=2.
- Solve 1/(x−2)=x−2.
Explained answers
1. x≠−1. 5=2(x+1), so x=3/2.
2. x≠0,−2. Multiply by x(x+2): (x+2)+x=x(x+2). So x²−2=0, giving x=±√2; both are allowed.
3. x≠4. Multiplying through gives x=4, which is excluded. No solution.
4. x≠1. x+2=2x−2, so x=4.
5. x≠2. Multiply through: 1=(x−2)². So x−2=±1, giving x=1 or 3.
Final thought
A fractional equation is often a polynomial equation wearing denominator structure. The denominator structure controls the domain; clearing it reveals the simpler equation underneath.
Record the restrictions first, clear the denominators cleanly, solve the hidden equation, then return to the original to decide which roots survive.
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