Question 1 – Solve pairs of equations by substitution method
📚 What is Substitution Method?
The substitution method is a technique for solving a system of linear equations. The steps are:
- Solve one equation for one variable in terms of the other
- Substitute this expression into the second equation
- Solve the resulting equation for the remaining variable
- Substitute back to find the other variable
- Verify the solution in both original equations
📑 Question Parts
Part (i) – Simple Linear Equations
Question: \( x + y = 14 \) and \( x – y = 4 \)
Step 1: Express one variable in terms of the other
From equation (1): \( x + y = 14 \)
\[ x = 14 – y \quad \text{…(3)} \]
Step 2: Substitute into the second equation
Substitute \( x = 14 – y \) in equation (2): \( x – y = 4 \)
\[ (14 – y) – y = 4 \]
\[ 14 – y – y = 4 \]
\[ 14 – 2y = 4 \]
\[ -2y = 4 – 14 \]
\[ -2y = -10 \]
\[ y = 5 \]
Step 3: Find the value of x
Substitute \( y = 5 \) in equation (3):
\[ x = 14 – 5 = 9 \]
Step 4: Verify the solution
Check in equation (1): \( x + y = 14 \)
LHS = \( 9 + 5 = 14 \) = RHS ✓
Check in equation (2): \( x – y = 4 \)LHS = \( 9 – 5 = 4 \) = RHS ✓
✅ Answer: \( x = 9 \) and \( y = 5 \)
Part (ii) – Equations with Fractions
Question: \( s – t = 3 \) and \( \frac{s}{3} + \frac{t}{2} = 6 \)
Step 1: Express one variable in terms of the other
From equation (1): \( s – t = 3 \)
\[ s = t + 3 \quad \text{…(3)} \]
Step 2: Substitute into the second equation
Substitute \( s = t + 3 \) in equation (2): \( \frac{s}{3} + \frac{t}{2} = 6 \)
\[ \frac{t + 3}{3} + \frac{t}{2} = 6 \]
Multiply throughout by 6 (LCM of 3 and 2):
\[ 6 \times \frac{t + 3}{3} + 6 \times \frac{t}{2} = 6 \times 6 \] \[ 2(t + 3) + 3t = 36 \] \[ 2t + 6 + 3t = 36 \] \[ 5t + 6 = 36 \] \[ 5t = 30 \] \[ t = 6 \]Step 3: Find the value of s
Substitute \( t = 6 \) in equation (3):
\[ s = 6 + 3 = 9 \]
Step 4: Verify the solution
Check in equation (1): \( s – t = 3 \)
LHS = \( 9 – 6 = 3 \) = RHS ✓
Check in equation (2): \( \frac{s}{3} + \frac{t}{2} = 6 \)LHS = \( \frac{9}{3} + \frac{6}{2} = 3 + 3 = 6 \) = RHS ✓
✅ Answer: \( s = 9 \) and \( t = 6 \)
Part (iii) – Coincident Lines (Special Case)
Question: \( 3x – y = 3 \) and \( 9x – 3y = 9 \)
Step 1: Check the relationship between equations
Observe equation (2):
\[ 9x – 3y = 9 \]
Divide throughout by 3:
\[ 3x – y = 3 \]This is exactly the same as equation (1)!
⚠️ Special Case: Coincident Lines
Both equations represent the same line. This means:
- The lines overlap completely
- There are infinitely many solutions
- Any point on the line \( 3x – y = 3 \) is a solution
Step 2: Express the solution
From \( 3x – y = 3 \):
\[ y = 3x – 3 \]
For any value of x, we can find y using this relation.
Step 3: Find some example solutions
Example solutions:
- If \( x = 0 \): \( y = 3(0) – 3 = -3 \) → Solution: (0, -3)
- If \( x = 1 \): \( y = 3(1) – 3 = 0 \) → Solution: (1, 0)
- If \( x = 2 \): \( y = 3(2) – 3 = 3 \) → Solution: (2, 3)
- If \( x = 3 \): \( y = 3(3) – 3 = 6 \) → Solution: (3, 6)
✅ Answer: Infinitely many solutions
General solution: \( y = 3x – 3 \) for any real value of x
General solution: \( y = 3x – 3 \) for any real value of x
Part (iv) – Equations with Decimals
Question: \( 0.2x + 0.3y = 1.3 \) and \( 0.4x + 0.5y = 2.3 \)
Step 1: Convert to whole numbers (optional but easier)
Multiply equation (1) by 10:
\[ 2x + 3y = 13 \quad \text{…(1′)} \]
Multiply equation (2) by 10:
\[ 4x + 5y = 23 \quad \text{…(2′)} \]
Step 2: Express one variable in terms of the other
From equation (1′): \( 2x + 3y = 13 \)
\[ 2x = 13 – 3y \]
\[ x = \frac{13 – 3y}{2} \quad \text{…(3)} \]
Step 3: Substitute into the second equation
Substitute \( x = \frac{13 – 3y}{2} \) in equation (2′): \( 4x + 5y = 23 \)
\[ 4 \times \frac{13 – 3y}{2} + 5y = 23 \]
\[ 2(13 – 3y) + 5y = 23 \]
\[ 26 – 6y + 5y = 23 \]
\[ 26 – y = 23 \]
\[ -y = 23 – 26 \]
\[ -y = -3 \]
\[ y = 3 \]
Step 4: Find the value of x
Substitute \( y = 3 \) in equation (3):
\[ x = \frac{13 – 3(3)}{2} = \frac{13 – 9}{2} = \frac{4}{2} = 2 \]
Step 5: Verify the solution
Check in original equation (1): \( 0.2x + 0.3y = 1.3 \)
LHS = \( 0.2(2) + 0.3(3) = 0.4 + 0.9 = 1.3 \) = RHS ✓
Check in original equation (2): \( 0.4x + 0.5y = 2.3 \)LHS = \( 0.4(2) + 0.5(3) = 0.8 + 1.5 = 2.3 \) = RHS ✓
✅ Answer: \( x = 2 \) and \( y = 3 \)
Part (v) – Equations with Surds
Question: \( \sqrt{2}x + \sqrt{3}y = 0 \) and \( \sqrt{3}x – \sqrt{8}y = 0 \)
Step 1: Express one variable in terms of the other
From equation (1): \( \sqrt{2}x + \sqrt{3}y = 0 \)
\[ \sqrt{2}x = -\sqrt{3}y \]
\[ x = \frac{-\sqrt{3}y}{\sqrt{2}} \]
\[ x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad \text{…(3)} \]
Step 2: Substitute into the second equation
Substitute \( x = -\frac{\sqrt{3}}{\sqrt{2}}y \) in equation (2): \( \sqrt{3}x – \sqrt{8}y = 0 \)
\[ \sqrt{3} \times \left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) – \sqrt{8}y = 0 \]
\[ -\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{2}}y – \sqrt{8}y = 0 \]
\[ -\frac{3}{\sqrt{2}}y – \sqrt{8}y = 0 \]
Note: \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
\[ -\frac{3}{\sqrt{2}}y – 2\sqrt{2}y = 0 \]Multiply throughout by \( \sqrt{2} \):
\[ -3y – 2\sqrt{2} \times \sqrt{2}y = 0 \] \[ -3y – 4y = 0 \] \[ -7y = 0 \] \[ y = 0 \]Step 3: Find the value of x
Substitute \( y = 0 \) in equation (3):
\[ x = -\frac{\sqrt{3}}{\sqrt{2}}(0) = 0 \]
Step 4: Verify the solution
Check in equation (1): \( \sqrt{2}x + \sqrt{3}y = 0 \)
LHS = \( \sqrt{2}(0) + \sqrt{3}(0) = 0 \) = RHS ✓
Check in equation (2): \( \sqrt{3}x – \sqrt{8}y = 0 \)LHS = \( \sqrt{3}(0) – \sqrt{8}(0) = 0 \) = RHS ✓
✅ Answer: \( x = 0 \) and \( y = 0 \)
Part (vi) – Complex Fractional Equations
Question: \( \frac{3x}{2} – \frac{5y}{3} = -2 \) and \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
Step 1: Clear fractions from both equations
For equation (1): \( \frac{3x}{2} – \frac{5y}{3} = -2 \)
Multiply throughout by 6 (LCM of 2 and 3):
\[ 6 \times \frac{3x}{2} – 6 \times \frac{5y}{3} = 6 \times (-2) \] \[ 9x – 10y = -12 \quad \text{…(1′)} \]
For equation (2): \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
Multiply throughout by 6 (LCM of 3, 2, and 6):
\[ 6 \times \frac{x}{3} + 6 \times \frac{y}{2} = 6 \times \frac{13}{6} \] \[ 2x + 3y = 13 \quad \text{…(2′)} \]Step 2: Express one variable in terms of the other
From equation (2′): \( 2x + 3y = 13 \)
\[ 2x = 13 – 3y \]
\[ x = \frac{13 – 3y}{2} \quad \text{…(3)} \]
Step 3: Substitute into the first equation
Substitute \( x = \frac{13 – 3y}{2} \) in equation (1′): \( 9x – 10y = -12 \)
\[ 9 \times \frac{13 – 3y}{2} – 10y = -12 \]
Multiply throughout by 2:
\[ 9(13 – 3y) – 20y = -24 \] \[ 117 – 27y – 20y = -24 \] \[ 117 – 47y = -24 \] \[ -47y = -24 – 117 \] \[ -47y = -141 \] \[ y = 3 \]Step 4: Find the value of x
Substitute \( y = 3 \) in equation (3):
\[ x = \frac{13 – 3(3)}{2} = \frac{13 – 9}{2} = \frac{4}{2} = 2 \]
Step 5: Verify the solution
Check in original equation (1): \( \frac{3x}{2} – \frac{5y}{3} = -2 \)
LHS = \( \frac{3(2)}{2} – \frac{5(3)}{3} = 3 – 5 = -2 \) = RHS ✓
Check in original equation (2): \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)LHS = \( \frac{2}{3} + \frac{3}{2} = \frac{4 + 9}{6} = \frac{13}{6} \) = RHS ✓
✅ Answer: \( x = 2 \) and \( y = 3 \)
📊 Summary of All Solutions
| Part | Equations | Solution |
|---|---|---|
| (i) | \( x + y = 14, x – y = 4 \) | \( x = 9, y = 5 \) |
| (ii) | \( s – t = 3, \frac{s}{3} + \frac{t}{2} = 6 \) | \( s = 9, t = 6 \) |
| (iii) | \( 3x – y = 3, 9x – 3y = 9 \) | Infinitely many solutions |
| (iv) | \( 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3 \) | \( x = 2, y = 3 \) |
| (v) | \( \sqrt{2}x + \sqrt{3}y = 0, \sqrt{3}x – \sqrt{8}y = 0 \) | \( x = 0, y = 0 \) |
| (vi) | \( \frac{3x}{2} – \frac{5y}{3} = -2, \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \) | \( x = 2, y = 3 \) |
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Forgetting to substitute back to find the second variable.
✅ Correct: After finding one variable, always substitute it back to find the other.
✅ Correct: After finding one variable, always substitute it back to find the other.
❌ Mistake 2: Making sign errors during substitution.
✅ Correct: Be careful with negative signs, especially when substituting expressions with brackets.
✅ Correct: Be careful with negative signs, especially when substituting expressions with brackets.
❌ Mistake 3: Not clearing fractions before solving.
✅ Correct: Multiply by LCM to eliminate fractions and make calculations easier.
✅ Correct: Multiply by LCM to eliminate fractions and make calculations easier.
❌ Mistake 4: Not verifying the solution in both original equations.
✅ Correct: Always check your answer in both equations to ensure it’s correct.
✅ Correct: Always check your answer in both equations to ensure it’s correct.
💡 Key Points to Remember
- Substitution Method: Express one variable in terms of the other and substitute
- Choose Wisely: Pick the equation and variable that’s easiest to isolate
- Clear Fractions: Multiply by LCM to eliminate fractions first
- Clear Decimals: Multiply by 10, 100, etc. to convert to whole numbers
- Simplify Surds: Rationalize and simplify square roots carefully
- Special Cases: Recognize coincident lines (infinitely many solutions)
- Always Verify: Check your solution in both original equations
- Sign Errors: Be extra careful with negative signs
- Bracket Expansion: Distribute correctly when substituting expressions
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.