Question 6 – Finding Conditions for Infinite/No Solutions
📚 Understanding Conditions for Solutions
For a pair of linear equations:
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
| Condition | Type of Solution | Lines |
|---|---|---|
| \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) | Unique solution | Intersecting |
| \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) | No solution | Parallel |
| \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) | Infinitely many solutions | Coincident |
Part (i) – For which values of a and b does the pair have infinite solutions?
\( 2x + 3y = 7 \)
\( (a – b)x + (a + b)y = 3a + b – 2 \)
Step 1: Write equations in standard form
Here: \( a_1 = 2, b_1 = 3, c_1 = -7 \)
Here: \( a_2 = (a – b), b_2 = (a + b), c_2 = -(3a + b – 2) \)
Step 2: Apply condition for infinite solutions
Condition for Infinite Solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]Step 3: Solve using first two ratios
Cross multiply:
\[ 2(a + b) = 3(a – b) \] \[ 2a + 2b = 3a – 3b \] \[ 2b + 3b = 3a – 2a \] \[ 5b = a \] \[ a = 5b \quad \text{…(3)} \]Step 4: Solve using first and third ratios
Cross multiply:
\[ 2(3a + b – 2) = 7(a – b) \] \[ 6a + 2b – 4 = 7a – 7b \] \[ 6a – 7a + 2b + 7b = 4 \] \[ -a + 9b = 4 \] \[ a = 9b – 4 \quad \text{…(4)} \]Step 5: Solve equations (3) and (4)
\( a = 5b \) and \( a = 9b – 4 \)
Therefore:
\[ 5b = 9b – 4 \] \[ 5b – 9b = -4 \] \[ -4b = -4 \] \[ b = 1 \]Step 6: Verification
Ratio 1: \( \frac{a_1}{a_2} = \frac{2}{a – b} = \frac{2}{5 – 1} = \frac{2}{4} = \frac{1}{2} \)
Ratio 2: \( \frac{b_1}{b_2} = \frac{3}{a + b} = \frac{3}{5 + 1} = \frac{3}{6} = \frac{1}{2} \)
Ratio 3: \( \frac{c_1}{c_2} = \frac{-7}{-(3a + b – 2)} = \frac{7}{3(5) + 1 – 2} = \frac{7}{14} = \frac{1}{2} \)
Since all three ratios are equal: \( \frac{1}{2} = \frac{1}{2} = \frac{1}{2} \) ✓
Equation (1): \( 2x + 3y = 7 \)
Equation (2): \( (5-1)x + (5+1)y = 3(5) + 1 – 2 \)
\( 4x + 6y = 14 \)
Dividing by 2: \( 2x + 3y = 7 \)
Both equations are identical! ✓
Visual Graph Representation
📊 Graphical Solution (Coincident Lines)
Graph showing coincident lines (infinite solutions) when a = 5, b = 1
Part (ii) – For which value of k will the pair have no solution?
\( 3x + y = 1 \)
\( (2k – 1)x + (k – 1)y = 2k + 1 \)
Step 1: Write equations in standard form
Here: \( a_1 = 3, b_1 = 1, c_1 = -1 \)
Here: \( a_2 = (2k – 1), b_2 = (k – 1), c_2 = -(2k + 1) \)
Step 2: Apply condition for no solution
Condition for No Solution (Parallel Lines):
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]Step 3: Solve for k
Cross multiply:
\[ 3(k – 1) = 1(2k – 1) \] \[ 3k – 3 = 2k – 1 \] \[ 3k – 2k = -1 + 3 \] \[ k = 2 \]Step 4: Verify that third ratio is different
Ratio 1: \( \frac{a_1}{a_2} = \frac{3}{2k – 1} = \frac{3}{2(2) – 1} = \frac{3}{3} = 1 \)
Ratio 2: \( \frac{b_1}{b_2} = \frac{1}{k – 1} = \frac{1}{2 – 1} = \frac{1}{1} = 1 \)
Ratio 3: \( \frac{c_1}{c_2} = \frac{-1}{-(2k + 1)} = \frac{1}{2(2) + 1} = \frac{1}{5} \)
Since: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = 1 \) but \( \frac{c_1}{c_2} = \frac{1}{5} \neq 1 \)
Condition satisfied! ✓
Equation (1): \( 3x + y = 1 \)
Equation (2) with k = 2: \( (2(2)-1)x + (2-1)y = 2(2) + 1 \)
\( 3x + y = 5 \)
Both have same slope but different intercepts → Parallel lines! ✓
Visual Graph Representation
📊 Graphical Solution (Parallel Lines)
Graph showing parallel lines (no solution) when k = 2
📊 Summary of Solutions
| Part | Condition | Values | Result |
|---|---|---|---|
| (i) | Infinite solutions (Coincident lines) | \( a = 5 \) \( b = 1 \) | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2} \) |
| (ii) | No solution (Parallel lines) | \( k = 2 \) | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = 1 \neq \frac{c_1}{c_2} = \frac{1}{5} \) |
⚠️ Common Mistakes to Avoid
✅ Correct:
- Infinite solutions: All three ratios equal
- No solution: First two ratios equal, third different
✅ Correct: Always verify all three ratios to confirm your answer.
✅ Correct: Be careful: \( ax + by = c \) becomes \( ax + by - c = 0 \), so constant term is \( -c \).
✅ Correct: When \( \frac{a}{b} = \frac{c}{d} \), then \( ad = bc \).
💡 Key Points to Remember
- Standard Form: Always write as \( ax + by + c = 0 \) (note the sign of c)
- Three Conditions:
- 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: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
- Cross Multiplication: Use when ratios are equal
- Verification: Always substitute back to verify all three ratios
- Geometric Interpretation:
- Infinite solutions → Lines overlap (coincident)
- No solution → Lines are parallel
- Unique solution → Lines intersect at one point
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.