Exercise 3.3 – Question 5: Equations with Reciprocals
Problem: Solve the following pair of equations by reducing them to a pair of linear equations:
\[ \frac{1}{2x} + \frac{1}{3y} = 2 \]
\[ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \]
📚 Understanding the Problem
Key Concept: These equations contain reciprocals of variables (\( \frac{1}{x} \) and \( \frac{1}{y} \)).
Strategy: Reduce to linear form by substitution:
- Let \( \frac{1}{x} = u \) (so \( x = \frac{1}{u} \))
- Let \( \frac{1}{y} = v \) (so \( y = \frac{1}{v} \))
- This converts the equations into linear form in terms of u and v
- Solve for u and v, then find x and y
📐 Method Overview
- Substitute: Replace \( \frac{1}{x} \) with u and \( \frac{1}{y} \) with v
- Simplify: Get linear equations in u and v
- Solve: Use elimination or substitution method
- Back-substitute: Find x and y from u and v
- Verify: Check in original equations
Step 1: Make Substitution
Let:
\( \frac{1}{x} = u \) …(A)
\( \frac{1}{y} = v \) …(B)
Note: This means \( x = \frac{1}{u} \) and \( y = \frac{1}{v} \)
Step 2: Rewrite Original Equations
Original equation (1): \( \frac{1}{2x} + \frac{1}{3y} = 2 \)
Rewrite using substitution:
\[ \frac{1}{2} \cdot \frac{1}{x} + \frac{1}{3} \cdot \frac{1}{y} = 2 \] \[ \frac{1}{2}u + \frac{1}{3}v = 2 \]Multiply by 6 (LCM of 2 and 3):
\[ 3u + 2v = 12 \quad \text{…(1)} \]
Original equation (2): \( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \)
Rewrite using substitution:
\[ \frac{1}{3} \cdot \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{y} = \frac{13}{6} \] \[ \frac{1}{3}u + \frac{1}{2}v = \frac{13}{6} \]Multiply by 6:
\[ 2u + 3v = 13 \quad \text{…(2)} \]Step 3: Solve Linear System
We now have a simple linear system:
Equation (1): \( 3u + 2v = 12 \)
Equation (2): \( 2u + 3v = 13 \)
Method: Elimination
Multiply equation (1) by 3 and equation (2) by 2:
Equation (1) × 3: \( 9u + 6v = 36 \) …(3)
Equation (2) × 2: \( 4u + 6v = 26 \) …(4)
Subtract equation (4) from equation (3):
\[ (9u + 6v) – (4u + 6v) = 36 – 26 \]
\[ 5u = 10 \]
\[ u = 2 \]
Substitute \( u = 2 \) in equation (1):
\[ 3(2) + 2v = 12 \]
\[ 6 + 2v = 12 \]
\[ 2v = 6 \]
\[ v = 3 \]
Step 4: Find x and y
Recall: \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \)
Since \( u = 2 \):
\[ \frac{1}{x} = 2 \] \[ x = \frac{1}{2} \]Since \( v = 3 \):
\[ \frac{1}{y} = 3 \] \[ y = \frac{1}{3} \]Step 5: Visual Representation
📊 Graphical Solution in u-v Plane
Graph showing the intersection of two linear equations in u and v
📊 Graph Interpretation:
- Blue Line: \( 3u + 2v = 12 \)
- Red Line: \( 2u + 3v = 13 \)
- Green Point: Intersection at (2, 3) representing \( u = 2, v = 3 \)
- Purple dashed lines: Show the coordinates of the solution
Step 6: Transformation Visualization
🔄 From (u, v) to (x, y)
Visual representation of the transformation from u-v space to x-y space
Step 7: Verification
Given: \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \)
Check in original equation (1): \( \frac{1}{2x} + \frac{1}{3y} = 2 \)
LHS = \( \frac{1}{2 \times \frac{1}{2}} + \frac{1}{3 \times \frac{1}{3}} \)
\[ = \frac{1}{1} + \frac{1}{1} \] \[ = 1 + 1 = 2 \]= RHS ✓
Check in original equation (2): \( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \)
LHS = \( \frac{1}{3 \times \frac{1}{2}} + \frac{1}{2 \times \frac{1}{3}} \)
\[ = \frac{1}{\frac{3}{2}} + \frac{1}{\frac{2}{3}} \] \[ = \frac{2}{3} + \frac{3}{2} \] \[ = \frac{4 + 9}{6} = \frac{13}{6} \]= RHS ✓
✅ Answer:
\( x = \frac{1}{2} \) and \( y = \frac{1}{3} \)
\( x = \frac{1}{2} \) and \( y = \frac{1}{3} \)
📊 Summary Table
| Step | Variable | Value |
|---|---|---|
| After substitution | \( u = \frac{1}{x} \) | 2 |
| \( v = \frac{1}{y} \) | 3 | |
| Final answer | \( x \) | \( \frac{1}{2} \) |
| \( y \) | \( \frac{1}{3} \) |
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Forgetting to substitute back to find x and y.
✅ Correct: After finding u and v, use \( x = \frac{1}{u} \) and \( y = \frac{1}{v} \)
✅ Correct: After finding u and v, use \( x = \frac{1}{u} \) and \( y = \frac{1}{v} \)
❌ Mistake 2: Confusing \( \frac{1}{2x} \) with \( \frac{1}{2} \times x \)
✅ Correct: \( \frac{1}{2x} = \frac{1}{2} \times \frac{1}{x} \)
✅ Correct: \( \frac{1}{2x} = \frac{1}{2} \times \frac{1}{x} \)
❌ Mistake 3: Not multiplying by LCM to clear fractions.
✅ Correct: Always multiply by LCM to simplify equations.
✅ Correct: Always multiply by LCM to simplify equations.
❌ Mistake 4: Verification errors with reciprocals.
✅ Correct: Remember: \( \frac{1}{\frac{a}{b}} = \frac{b}{a} \)
✅ Correct: Remember: \( \frac{1}{\frac{a}{b}} = \frac{b}{a} \)
💡 Key Points to Remember
Solving Equations with Reciprocals
- Substitution Strategy:
- Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)
- This converts non-linear equations to linear form
- Solve for u and v first
- Then find x and y using \( x = \frac{1}{u} \) and \( y = \frac{1}{v} \)
- Working with Fractions:
- \( \frac{1}{2x} = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2}u \)
- Always multiply by LCM to clear denominators
- Simplify before solving
- Verification:
- Substitute x and y back into original equations
- Be careful with reciprocal calculations
- \( \frac{1}{\frac{1}{2}} = 2 \), not \( \frac{1}{2} \)
- Graphical Understanding:
- In u-v space: equations are linear (straight lines)
- Solution is the intersection point
- Transform back to x-y space using reciprocals
🎉 CONGRATULATIONS! 🎉
You've completed ALL 5 questions of Exercise 3.3!
Chapter 3: Pair of Linear Equations in Two Variables
You've completed ALL 5 questions of Exercise 3.3!
Chapter 3: Pair of Linear Equations in Two Variables
📚 Exercise 3.3 Complete Summary
All Questions Solved ✅
| Question | Topic | Method |
|---|---|---|
| Q1 | 6 algebraic problems | Substitution |
| Q2 | 7 word problems | Substitution |
| Q3 | 3 word problems | Elimination |
| Q4 | Boat and stream | Elimination |
| Q5 | Reciprocal equations | Substitution + Elimination |
📚 Related Questions
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.