Pair of Linear Equations in Two Variables — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Pair of Linear Equations in Two Variables" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Pair of Linear Equations in Two Variables" — 9 important questions with detailed a…
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NCERT Solutions for Class 10 Mathematics — Pair of Linear Equations in Two Variables. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Pair of Linear Equations in Two Variables — All 9 Questions Solved
Q1. Define a linear equation in two variables and state the general form.
Linear Equation in Two Variables:
A linear equation in two variables is an equation of the form ax + by + c = 0, where a, b, and c are real numbers and at least one of a or b is non-zero.
General form: ax + by + c = 0
or ax + by = c
where a = coefficient of x
b = coefficient of y
c = constant term
Examples:
- 2x + 3y - 6 = 0
- x - 2y = 5
- 3x + 4y = 12
Properties:
- The graph of a linear equation in two variables is a straight line
- The equation has infinitely many solutions (ordered pairs)
- Each solution corresponds to a point on the line
Q2. A pair of linear equations is consistent if which condition is satisfied?
Consistency of a Pair of Linear Equations:
Consider the pair: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
The pair of equations is CONSISTENT if it has at least one solution.
Two cases of consistency:
1) Consistent and Independent:
Condition: a₁/a₂ ≠ b₁/b₂
Number of solutions: Exactly one (unique)
Geometrically: The lines intersect at one point
2) Consistent and Dependent:
Condition: a₁/a₂ = b₁/b₂ = c₁/c₂
Number of solutions: Infinitely many
Geometrically: The lines coincide (are the same line)
Inconsistent:
Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Number of solutions: None
Geometrically: The lines are parallel and never intersect
Q3. Solve the system of equations 2x + y = 7 and x - y = 2 using the elimination method.
Given equations:
2x + y = 7 ... (1)
x - y = 2 ... (2)
Step 1: Add equations (1) and (2) to eliminate y
(2x + y) + (x - y) = 7 + 2
3x = 9
x = 3
Step 2: Substitute x = 3 into equation (2)
3 - y = 2
y = 1
Step 3: Verification
Substitute x = 3, y = 1 into equation (1):
2(3) + 1 = 6 + 1 = 7 ✓
Substitute x = 3, y = 1 into equation (2):
3 - 1 = 2 ✓
Therefore, the solution is x = 3, y = 1
Q4. Solve the system 3x + 4y = 10 and 2x - 2y = 2 using substitution method.
Given equations:
3x + 4y = 10 ... (1)
2x - 2y = 2 ... (2)
Step 1: Simplify equation (2) by dividing by 2
x - y = 1
x = y + 1 ... (3)
Step 2: Substitute x = y + 1 into equation (1)
3(y + 1) + 4y = 10
3y + 3 + 4y = 10
7y + 3 = 10
7y = 7
y = 1
Step 3: Find x using equation (3)
x = 1 + 1 = 2
Step 4: Verification
Substitute x = 2, y = 1 into equation (1):
3(2) + 4(1) = 6 + 4 = 10 ✓
Substitute x = 2, y = 1 into equation (2):
2(2) - 2(1) = 4 - 2 = 2 ✓
Therefore, the solution is x = 2, y = 1
Q5. Solve using the cross-multiplication method: x + 2y = 5 and 2x + 3y = 8
Given equations:
x + 2y = 5 ... (1)
2x + 3y = 8 ... (2)
Rearranging in the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
x + 2y - 5 = 0
2x + 3y - 8 = 0
Here: a₁ = 1, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = 3, c₂ = -8
Using cross-multiplication method:
x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
Calculate each denominator:
b₁c₂ - b₂c₁ = 2(-8) - 3(-5) = -16 + 15 = -1
c₁a₂ - c₂a₁ = (-5)(2) - (-8)(1) = -10 + 8 = -2
a₁b₂ - a₂b₁ = 1(3) - 2(2) = 3 - 4 = -1
So: x/(-1) = y/(-2) = 1/(-1)
Therefore: x = -1/(-1) = 1
y = -2/(-1) = 2
Verification:
1 + 2(2) = 1 + 4 = 5 ✓
2(1) + 3(2) = 2 + 6 = 8 ✓
Therefore, the solution is x = 1, y = 2
Q6. For what value of k does the system 2x - 3y = 7 and 4x - ky = 14 have infinitely many solutions?
For a system of linear equations to have infinitely many solutions, the equations must be consistent and dependent.
This means: a₁/a₂ = b₁/b₂ = c₁/c₂
For the given equations:
2x - 3y = 7 → a₁ = 2, b₁ = -3, c₁ = 7
4x - ky = 14 → a₂ = 4, b₂ = -k, c₂ = 14
Applying the condition:
a₁/a₂ = b₁/b₂ = c₁/c₂
First, check c₁/c₂:
7/14 = 1/2
So we need a₁/a₂ = 1/2:
2/4 = 1/2 ✓ (This is satisfied)
Now, we need b₁/b₂ = 1/2:
(-3)/(-k) = 1/2
3/k = 1/2
k = 6
Verification: With k = 6:
The second equation becomes 4x - 6y = 14, which simplifies to 2x - 3y = 7 (dividing by 2).
The equations are identical, so there are infinitely many solutions.
Therefore, k = 6
Q7. Determine if the following system is consistent: 3x + 2y = 5 and 6x + 4y = 8
For the system:
3x + 2y = 5 ... (1)
6x + 4y = 8 ... (2)
Identify coefficients:
a₁ = 3, b₁ = 2, c₁ = 5
a₂ = 6, b₂ = 4, c₂ = 8
Check the ratios:
a₁/a₂ = 3/6 = 1/2
b₁/b₂ = 2/4 = 1/2
c₁/c₂ = 5/8
Observation:
a₁/a₂ = b₁/b₂ = 1/2, but c₁/c₂ = 5/8 ≠ 1/2
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system is INCONSISTENT.
Geometrically: The lines represented by these equations are parallel and do not intersect.
To verify, simplify equation (2):
6x + 4y = 8
Dividing by 2: 3x + 2y = 4
Now we have:
3x + 2y = 5
3x + 2y = 4
These are two different parallel lines with the same slope but different y-intercepts. They never intersect, so the system has NO SOLUTION.
Therefore, the system is inconsistent
Q8. The sum of two numbers is 15 and their difference is 3. Find the numbers.
Let the two numbers be x and y.
Step 1: Form equations from the given conditions
Sum = 15: x + y = 15 ... (1)
Difference = 3: x - y = 3 ... (2)
Step 2: Solve using elimination method
Add equations (1) and (2):
(x + y) + (x - y) = 15 + 3
2x = 18
x = 9
Step 3: Find y
Substitute x = 9 into equation (1):
9 + y = 15
y = 6
Step 4: Verification
Sum: 9 + 6 = 15 ✓
Difference: 9 - 6 = 3 ✓
Therefore, the two numbers are 9 and 6
Q9. A train travels 300 km at a uniform speed. If the speed were increased by 5 km/h, the journey would take 2 hours less. Find the original speed of the train.
Let the original speed of the train be x km/h.
Step 1: Form equations using distance = speed × time
For original journey: Time = 300/x hours
For increased speed journey: Time = 300/(x + 5) hours
Given that the faster journey takes 2 hours less:
300/x - 300/(x + 5) = 2
Step 2: Solve the equation
300(x + 5) - 300x = 2x(x + 5)
300x + 1500 - 300x = 2x² + 10x
1500 = 2x² + 10x
2x² + 10x - 1500 = 0
Divide by 2:
x² + 5x - 750 = 0
Step 3: Factorise
Find two numbers that multiply to -750 and add to 5.
Factors: 30 and -25 (since 30 × (-25) = -750 and 30 + (-25) = 5)
x² + 5x - 750 = (x + 30)(x - 25) = 0
Step 4: Find the roots
Either x + 30 = 0 or x - 25 = 0
Therefore: x = -30 or x = 25
Since speed cannot be negative, x = 25 km/h.
Step 5: Verification
Original time: 300/25 = 12 hours
Time at increased speed (30 km/h): 300/30 = 10 hours
Difference: 12 - 10 = 2 hours ✓
Therefore, the original speed of the train is 25 km/h
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.