Question 3 – Check consistency of linear equations by comparing ratios
📑 Question Parts
📚 Understanding Consistency
A pair of linear equations is:
- Consistent: If it has at least one solution (either unique or infinitely many)
- Inconsistent: If it has no solution (parallel lines)
Using Ratio Method:
For equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):
- If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) → Consistent (Unique solution)
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) → Consistent (Infinitely many solutions)
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) → Inconsistent (No solution)
Part (i)
\( 3x + 2y = 5 \)
\( 2x – 3y = 7 \)
Solution Steps
Rewrite equations in the form \( ax + by + c = 0 \):
\( 3x + 2y – 5 = 0 \)
\( 2x – 3y – 7 = 0 \)
For equation \( 3x + 2y – 5 = 0 \):
\( a_1 = 3 \), \( b_1 = 2 \), \( c_1 = -5 \)
For equation \( 2x – 3y – 7 = 0 \):
\( a_2 = 2 \), \( b_2 = -3 \), \( c_2 = -7 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{3}{2} \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{2}{-3} = \frac{-2}{3} \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7} \]\( \frac{3}{2} = 1.5 \) and \( \frac{-2}{3} \approx -0.667 \)
Clearly, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Therefore, the lines intersect at exactly one point.
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the pair of equations has a unique solution.
The pair is CONSISTENT.
Part (ii)
\( 4x – 6y = 9 \)
Solution Steps
\( 2x – 3y – 8 = 0 \)
\( 4x – 6y – 9 = 0 \)
For equation \( 2x – 3y – 8 = 0 \):
\( a_1 = 2 \), \( b_1 = -3 \), \( c_1 = -8 \)
For equation \( 4x – 6y – 9 = 0 \):
\( a_2 = 4 \), \( b_2 = -6 \), \( c_2 = -9 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2} \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9} \]\( \frac{a_1}{a_2} = \frac{1}{2} \) and \( \frac{b_1}{b_2} = \frac{1}{2} \)
But \( \frac{c_1}{c_2} = \frac{8}{9} \approx 0.889 \)
Therefore: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
This means the lines are parallel.
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel and never intersect.
The pair has no solution and is INCONSISTENT.
Part (iii)
\( 9x – 10y = 14 \)
Solution Steps
Multiply \( \frac{3}{2}x + \frac{5}{3}y = 7 \) by 6 (LCM of 2 and 3):
\[ 6 \times \frac{3}{2}x + 6 \times \frac{5}{3}y = 6 \times 7 \] \[ 9x + 10y = 42 \] \[ 9x + 10y – 42 = 0 \]\( 9x + 10y – 42 = 0 \)
\( 9x – 10y – 14 = 0 \)
For equation \( 9x + 10y – 42 = 0 \):
\( a_1 = 9 \), \( b_1 = 10 \), \( c_1 = -42 \)
For equation \( 9x – 10y – 14 = 0 \):
\( a_2 = 9 \), \( b_2 = -10 \), \( c_2 = -14 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{9}{9} = 1 \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{10}{-10} = -1 \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-42}{-14} = 3 \]\( \frac{a_1}{a_2} = 1 \) and \( \frac{b_1}{b_2} = -1 \)
Clearly, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Therefore, the lines intersect at one point.
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the pair has a unique solution.
The pair is CONSISTENT.
Part (iv)
\( -10x + 6y = -22 \)
Solution Steps
\( 5x – 3y – 11 = 0 \)
\( -10x + 6y + 22 = 0 \)
For equation \( 5x – 3y – 11 = 0 \):
\( a_1 = 5 \), \( b_1 = -3 \), \( c_1 = -11 \)
For equation \( -10x + 6y + 22 = 0 \):
\( a_2 = -10 \), \( b_2 = 6 \), \( c_2 = 22 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{5}{-10} = \frac{-1}{2} \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{-3}{6} = \frac{-1}{2} \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-11}{22} = \frac{-1}{2} \]We observe that:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{-1}{2} \]All three ratios are equal!
If we multiply the first equation by -2:
\[ -2(5x – 3y – 11) = 0 \] \[ -10x + 6y + 22 = 0 \]This is exactly the second equation! Both represent the same line.
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident.
The pair has infinitely many solutions and is CONSISTENT.
Part (v)
\( 2x + 3y = 12 \)
Solution Steps
Multiply \( \frac{4}{3}x + 2y = 8 \) by 3:
\[ 3 \times \frac{4}{3}x + 3 \times 2y = 3 \times 8 \] \[ 4x + 6y = 24 \] \[ 4x + 6y – 24 = 0 \]\( 4x + 6y – 24 = 0 \)
\( 2x + 3y – 12 = 0 \)
For equation \( 4x + 6y – 24 = 0 \):
\( a_1 = 4 \), \( b_1 = 6 \), \( c_1 = -24 \)
For equation \( 2x + 3y – 12 = 0 \):
\( a_2 = 2 \), \( b_2 = 3 \), \( c_2 = -12 \)
Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{4}{2} = 2 \]Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{6}{3} = 2 \]Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-24}{-12} = 2 \]We observe that:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 2 \]All three ratios are equal!
If we divide the first equation by 2:
\[ \frac{1}{2}(4x + 6y – 24) = 0 \] \[ 2x + 3y – 12 = 0 \]This is exactly the second equation! Both represent the same line.
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident.
The pair has infinitely many solutions and is CONSISTENT.
📊 Summary of All Five Parts
| Part | \( \frac{a_1}{a_2} \) | \( \frac{b_1}{b_2} \) | \( \frac{c_1}{c_2} \) | Condition | Consistency | Solution Type |
|---|---|---|---|---|---|---|
| (i) | \( \frac{3}{2} \) | \( \frac{-2}{3} \) | \( \frac{5}{7} \) | \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) | Consistent | Unique |
| (ii) | \( \frac{1}{2} \) | \( \frac{1}{2} \) | \( \frac{8}{9} \) | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) | Inconsistent | No solution |
| (iii) | 1 | -1 | 3 | \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) | Consistent | Unique |
| (iv) | \( \frac{-1}{2} \) | \( \frac{-1}{2} \) | \( \frac{-1}{2} \) | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) | Consistent | Infinitely many |
| (v) | 2 | 2 | 2 | \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) | Consistent | Infinitely many |
⚠️ Common Mistakes to Avoid
✅ Correct: Always multiply by the LCM to clear fractions before identifying coefficients.
✅ Correct: Consistent means at least one solution exists. Inconsistent means no solution.
✅ Correct: For \( 2x – 3y = 8 \), the standard form is \( 2x – 3y – 8 = 0 \), so \( c_1 = -8 \).
✅ Correct: Always reduce fractions to lowest terms for accurate comparison.
💡 Key Points to Remember
- Consistent System: Has at least one solution (unique or infinitely many)
- Inconsistent System: Has no solution (parallel lines)
- Three Conditions:
- \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) → Consistent (Unique)
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) → Consistent (Infinitely many)
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) → Inconsistent (No solution)
- Standard Form: Always convert to \( ax + by + c = 0 \) first
- Fraction Handling: Multiply by LCM to eliminate fractions
- Sign Awareness: Include negative signs in coefficients
- Verification: For coincident lines, one equation should be a multiple of the other
- Quick Check: Compare only first two ratios initially; if equal, check third ratio
📝 Practice Similar Problems
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.