Conditions for Solvability
A pair of linear equations in two variables may have a unique solution, infinitely many solutions, or no solution. The type of solution is determined by comparing the ratios of the coefficients.
Before solving a system of equations, it is useful to determine whether a solution exists and what kind it is. This saves time and avoids attempting to solve inconsistent systems.
The conditions for solvability are derived from the algebraic relationship between the coefficients of the two equations and correspond to the graphical behaviour of the two lines.
What is Conditions for Solvability of Pair of Linear Equations?
Definition: Given a pair of linear equations:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
The conditions for solvability are determined by the ratios a₁/a₂, b₁/b₂, and c₁/c₂.
Three cases:
| Condition | Type of Solution | Graphical Meaning | Consistency |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Unique solution (one point) | Lines intersect at one point | Consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Infinitely many solutions | Lines are coincident (same line) | Consistent (dependent) |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | No solution | Lines are parallel | Inconsistent |
Conditions for Solvability Formula
Conditions at a Glance:
Unique: a₁/a₂ ≠ b₁/b₂
Infinite: a₁/a₂ = b₁/b₂ = c₁/c₂
None: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Where:
- a₁, b₁, c₁ = coefficients of the first equation
- a₂, b₂, c₂ = coefficients of the second equation
- Both equations must be in the form ax + by + c = 0 before comparing ratios.
Important: Ensure the constant term is on the correct side. If the equations are given as a₁x + b₁y = d₁, rewrite as a₁x + b₁y + (−d₁) = 0, so c₁ = −d₁.
Derivation and Proof
Why do these conditions work?
- Two linear equations represent two straight lines in the xy-plane.
- Two distinct lines can: (a) intersect at one point, (b) be parallel, or (c) be the same line.
- The equation a₁x + b₁y + c₁ = 0 has slope = −a₁/b₁ and y-intercept = −c₁/b₁.
- The equation a₂x + b₂y + c₂ = 0 has slope = −a₂/b₂ and y-intercept = −c₂/b₂.
- Different slopes (−a₁/b₁ ≠ −a₂/b₂) means a₁/a₂ ≠ b₁/b₂ → lines intersect → unique solution.
- Same slope, different intercept means a₁/a₂ = b₁/b₂ but c₁/c₂ differs → lines are parallel → no solution.
- Same slope, same intercept means a₁/a₂ = b₁/b₂ = c₁/c₂ → same line → infinitely many solutions.
Types and Properties
Types of systems based on consistency:
- Consistent system: Has at least one solution. Includes both unique-solution and infinite-solution cases.
- Inconsistent system: Has no solution. The equations contradict each other.
- Dependent system: A special case of consistent — the two equations are equivalent (one is a multiple of the other).
Graphical interpretation:
- Intersecting lines: Unique solution. The point of intersection gives x and y.
- Parallel lines: No solution. The lines never meet, so no (x, y) satisfies both.
- Coincident lines: Infinitely many solutions. Every point on the line satisfies both equations.
Methods
Steps to determine the type of solution:
- Write both equations in standard form: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
- Identify coefficients: a₁, b₁, c₁, a₂, b₂, c₂.
- Calculate the three ratios: a₁/a₂, b₁/b₂, c₁/c₂.
- Compare:
- If a₁/a₂ ≠ b₁/b₂ → unique solution.
- If a₁/a₂ = b₁/b₂ = c₁/c₂ → infinite solutions.
- If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution.
Common mistakes:
- Not converting to standard form (ax + by + c = 0) before comparing. If the equation is 2x + 3y = 5, then c = −5, not 5.
- Comparing absolute values instead of actual ratios (ignoring signs).
- Confusing "consistent" with "unique solution" — consistent includes infinite solutions too.
Solved Examples
Example 1: Unique Solution (Intersecting Lines)
Problem: Determine the type of solution for: 2x + 3y = 5 and 4x − y = 7.
Solution:
Standard form:
- 2x + 3y − 5 = 0 → a₁ = 2, b₁ = 3, c₁ = −5
- 4x − y − 7 = 0 → a₂ = 4, b₂ = −1, c₂ = −7
Ratios:
- a₁/a₂ = 2/4 = 1/2
- b₁/b₂ = 3/(−1) = −3
- a₁/a₂ ≠ b₁/b₂ (1/2 ≠ −3)
Answer: Unique solution (lines intersect). System is consistent.
Example 2: No Solution (Parallel Lines)
Problem: Determine the type of solution for: 3x + 2y = 8 and 6x + 4y = 10.
Solution:
Standard form:
- 3x + 2y − 8 = 0 → a₁ = 3, b₁ = 2, c₁ = −8
- 6x + 4y − 10 = 0 → a₂ = 6, b₂ = 4, c₂ = −10
Ratios:
- a₁/a₂ = 3/6 = 1/2
- b₁/b₂ = 2/4 = 1/2
- c₁/c₂ = (−8)/(−10) = 4/5
- a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (1/2 = 1/2 ≠ 4/5)
Answer: No solution (parallel lines). System is inconsistent.
Example 3: Infinitely Many Solutions (Coincident Lines)
Problem: Determine the type of solution for: 2x + 3y = 6 and 4x + 6y = 12.
Solution:
Standard form:
- 2x + 3y − 6 = 0 → a₁ = 2, b₁ = 3, c₁ = −6
- 4x + 6y − 12 = 0 → a₂ = 4, b₂ = 6, c₂ = −12
Ratios:
- a₁/a₂ = 2/4 = 1/2
- b₁/b₂ = 3/6 = 1/2
- c₁/c₂ = (−6)/(−12) = 1/2
- a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2
Answer: Infinitely many solutions (coincident lines). System is consistent and dependent.
Example 4: Finding k for Unique Solution
Problem: For what value of k does the system kx + 3y = k − 3 and 12x + ky = k have a unique solution?
Solution:
Standard form:
- kx + 3y − (k−3) = 0 → a₁ = k, b₁ = 3
- 12x + ky − k = 0 → a₂ = 12, b₂ = k
For unique solution: a₁/a₂ ≠ b₁/b₂
- k/12 ≠ 3/k
- k² ≠ 36
- k ≠ ±6
Answer: The system has a unique solution for all values of k except k = 6 and k = −6.
Example 5: Finding k for No Solution
Problem: Find the value of k for which the system 2x + ky = 10 and 3x + (k+1)y = 12 has no solution.
Solution:
Standard form:
- a₁ = 2, b₁ = k, c₁ = −10
- a₂ = 3, b₂ = k+1, c₂ = −12
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
- 2/3 = k/(k+1)
- 2(k+1) = 3k
- 2k + 2 = 3k
- k = 2
Verify c₁/c₂ ≠ a₁/a₂:
- c₁/c₂ = (−10)/(−12) = 5/6
- a₁/a₂ = 2/3
- 5/6 ≠ 2/3 ✔
Answer: k = 2
Example 6: Finding k for Infinite Solutions
Problem: Find the value of k for which the system (k−1)x + 3y = 7 and (k+1)x + 6y = 14 has infinitely many solutions.
Solution:
Coefficients:
- a₁ = k−1, b₁ = 3, c₁ = −7
- a₂ = k+1, b₂ = 6, c₂ = −14
For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
- b₁/b₂ = 3/6 = 1/2
- c₁/c₂ = (−7)/(−14) = 1/2 ✔
- a₁/a₂ = (k−1)/(k+1) = 1/2
- 2(k−1) = k+1
- 2k − 2 = k + 1
- k = 3
Answer: k = 3
Example 7: Determining Consistency
Problem: Check whether the following system is consistent: x + 2y = 4, 2x + 4y = 12.
Solution:
Standard form:
- a₁ = 1, b₁ = 2, c₁ = −4
- a₂ = 2, b₂ = 4, c₂ = −12
Ratios:
- a₁/a₂ = 1/2
- b₁/b₂ = 2/4 = 1/2
- c₁/c₂ = (−4)/(−12) = 1/3
- a₁/a₂ = b₁/b₂ = 1/2 but c₁/c₂ = 1/3
- 1/2 ≠ 1/3
Answer: The system is inconsistent (no solution). The lines are parallel.
Example 8: Finding Two Values of k
Problem: Find the values of k for which the system 5x + 2y = k and 10x + 4y = 3 has (a) infinite solutions, (b) no solution.
Solution:
Coefficients:
- a₁ = 5, b₁ = 2, c₁ = −k
- a₂ = 10, b₂ = 4, c₂ = −3
Ratios:
- a₁/a₂ = 5/10 = 1/2
- b₁/b₂ = 2/4 = 1/2
- c₁/c₂ = (−k)/(−3) = k/3
(a) For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
- k/3 = 1/2 → k = 3/2
(b) For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
- k/3 ≠ 1/2 → k ≠ 3/2
Answer: (a) k = 3/2 for infinite solutions. (b) k ≠ 3/2 (any other value) for no solution. Note: unique solution is not possible since a₁/a₂ = b₁/b₂ always.
Real-World Applications
Business Planning:
- Checking whether supply and demand equations have a feasible equilibrium point (unique solution) or represent contradictory constraints (no solution).
Engineering:
- Determining if a system of force equations has a unique equilibrium or is indeterminate.
Network Analysis:
- In circuit analysis (Kirchhoff's laws), checking whether the system of equations for current has a unique solution.
CBSE Board Exams:
- Questions frequently ask: "For what value of k does the system have no solution / unique solution / infinitely many solutions?" This directly tests these conditions.
Key Points to Remember
- For unique solution: a₁/a₂ ≠ b₁/b₂ (lines intersect).
- For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂ (coincident lines).
- For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (parallel lines).
- Always write equations in standard form ax + by + c = 0 before comparing ratios.
- A system with at least one solution is called consistent.
- A system with no solution is called inconsistent.
- If the second equation is a scalar multiple of the first, the lines are coincident.
- These conditions are equivalent to checking whether the slopes and y-intercepts of the two lines match.
- "Find k for which..." problems are solved by setting up ratio equations and solving for k.
- This topic is essential for CBSE Class 10 — regularly appears as 2–3 mark questions.
Practice Problems
- Check whether the system 3x + 4y = 7 and 6x − 2y = 9 has a unique solution, no solution, or infinite solutions.
- For what value of p does the system px + 3y = p − 3, 12x + py = p have no solution?
- Find the value of k for which 2x + ky = 1, 3x − 5y = 7 has a unique solution.
- Determine the consistency of: 5x − 3y = 11, −10x + 6y = −22.
- For what value of a does the pair ax + 3y = a − 2, 12x + ay = a have infinitely many solutions?
- Show that the system 4x − 5y = 3, 8x − 10y = 1 is inconsistent.
Frequently Asked Questions
Q1. What are the conditions for a unique solution?
For the system a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, a unique solution exists when a₁/a₂ ≠ b₁/b₂. This means the two lines have different slopes and must intersect at exactly one point.
Q2. When does a pair of linear equations have no solution?
No solution exists when a₁/a₂ = b₁/b₂ ≠ c₁/c₂. The lines have the same slope but different y-intercepts, so they are parallel and never intersect.
Q3. What does 'consistent' mean?
A system is consistent if it has at least one solution. This includes both unique-solution (intersecting lines) and infinite-solution (coincident lines) cases. A system is inconsistent if it has no solution (parallel lines).
Q4. Do I need to write equations in standard form first?
Yes. The conditions use coefficients from ax + by + c = 0. If the equation is given as 2x + 3y = 5, rewrite it as 2x + 3y + (−5) = 0, so c = −5. Getting the sign of c wrong leads to wrong conclusions.
Q5. What if one equation is a multiple of the other?
Then all three ratios a₁/a₂ = b₁/b₂ = c₁/c₂ are equal. The equations represent the same line, and there are infinitely many solutions. Example: 2x + 3y = 6 and 4x + 6y = 12.
Q6. Can I use these conditions for non-linear equations?
No. These conditions apply only to linear equations in two variables. For non-linear systems (quadratic, etc.), you must solve them directly to determine the number of solutions.
Q7. What is a dependent system?
A dependent system is one where the two equations represent the same line (coincident). It has infinitely many solutions. Every dependent system is consistent, but not every consistent system is dependent.
Q8. How does this relate to graphical methods?
The ratio conditions predict the graphical outcome without drawing graphs. a₁/a₂ ≠ b₁/b₂ means intersecting lines, a₁/a₂ = b₁/b₂ = c₁/c₂ means coincident lines, and a₁/a₂ = b₁/b₂ ≠ c₁/c₂ means parallel lines.
Q9. How is this topic tested in CBSE exams?
Common question types: (1) Determine the type of solution for a given pair. (2) Find the value of k for which the system has unique/no/infinite solutions. (3) Check consistency of a given system. These carry 2–3 marks.
Q10. What happens if both a₁ and a₂ are zero?
If both a-coefficients are zero, the equations become b₁y + c₁ = 0 and b₂y + c₂ = 0. These are equations in y only. If b₁/b₂ = c₁/c₂, they are the same equation (infinite solutions for y, x can be anything). If not, they give different values of y — no solution.
Related Topics
- Consistency of Linear Equations
- Pair of Linear Equations in Two Variables
- Graphical Method for Linear Equations
- Substitution Method
- Elimination Method
- Cross-Multiplication Method
- Word Problems on Pair of Linear Equations
- Word Problems Solved Graphically
- Reducing Equations to Linear Form
- Linear Equations in Real Life
