Pair of Linear Equations in Two Variables Class 10 Previous Year Questions (Chapter 3)

1. 1

   Step 1: Points of x + 3y = 6

   1 mark

   Line 1: x + 3y = 6. The points (0, 2) and (6, 0) lie on this line.

   Why? Two points determine a straight line, and both points satisfy x + 3y = 6.

2. 2

   Step 2: Points of 2x - 3y = 12

   1 mark

   Line 2: 2x - 3y = 12. The points (0, -4) and (3, -2) lie on this line.

   Why? Two points determine a straight line, and both points satisfy 2x - 3y = 12.

3. 3

   Step 3: Graph and point of intersection

   1 mark

   Plot (0, 2), (6, 0), (0, -4), (3, -2). Join (0, 2) to (6, 0) and (0, -4) to (3, -2). The lines meet at (6, 0), so x = 6, y = 0.

   Why? When two lines intersect in a single point, that point is the unique common solution of the pair, so the pair is consistent.

Marks: 1 + 1 + 1 = 3 · Split as in the CBSE marking scheme, 2025 · Basic 430/1/1 Q28 (the marks split is taken from the 2025 Basic paper, where this question also appeared)

1. 1

   Step 1: Write the condition for infinitely many solutions

   1 mark (suggested)

   For infinitely many solutions, the lines must be coincident, so a₁/a₂ = b₁/b₂ = c₁/c₂.

   Why? For a dependent pair of linear equations, the two lines are coincident, so the ratios of the x-coefficients, y-coefficients and constant terms must all be equal.

2. 2

   Step 2: Solve the ratios and choose c

   1 mark (suggested)

   Since c is in a denominator, c ≠ 0. The ratios c/12 and 3/c are equal, so c² = 36, giving c = 6 or c = −6. The third ratio must also match: for c = 6, (3−c)/(−c) = 1/2 = c/12; for c = −6, (3−c)/(−c) = 3/2 ≠ c/12 = −1/2. Therefore, c = 6.

   Why? Every ratio in the condition must be equal, so a candidate found from the first equality must also satisfy the third ratio; this removes the candidate that does not.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Add the equations

   1 mark

   Adding the two equations: (31x + 43y) + (43x + 31y) = 117 + 105, i.e., 74x + 74y = 222, so x + y = 3.

   Why? Adding equal quantities on both sides of two equations gives another true equation; this is the first move of the elimination method.

2. 2

   Step 2: Subtract the equations

   1 mark

   Subtracting the first equation from the second: (43x + 31y) − (31x + 43y) = 105 − 117, i.e., 12x − 12y = −12, so x − y = −1.

   Why? Subtracting equal quantities on both sides keeps the equality true; the interchange of coefficients leaves only x and y terms with opposite signs.

3. 3

   Step 3: Solve for x and y

   1 mark

   Adding x + y = 3 and x − y = −1 gives 2x = 2, so x = 1. Then y = 3 − 1 = 2. Hence, the solution is x = 1 and y = 2.

   Why? Adding the two new equations eliminates y and gives x; substituting x back into x + y = 3 gives y.

Marks: 1 + 1 + 1 = 3 · Split as in the CBSE marking scheme, 2026 · Basic 430(B) Q27

1. 1

   Step 1: Form the equations

   1 mark

   Let the monthly incomes be ₹9x and ₹7x, and the monthly expenditures be ₹4y and ₹3y. Since savings = income − expenditure and each saves ₹5000, the equations are 9x − 4y = 5000 and 7x − 3y = 5000.

   Why? A ratio a : b is written as the multiples ax and bx, and savings = income − expenditure turns each person's condition into a linear equation.

2. 2

   Step 2: Eliminate y

   1 mark

   Multiply the first equation by 3 and the second by 4: 27x − 12y = 15000 and 28x − 12y = 20000. Subtract: (28x − 27x) = 20000 − 15000, so x = 5000.

   Why? The elimination method makes the coefficients of y equal, so subtracting the equations cancels y and leaves x.

3. 3

   Step 3: Find y

   ½ mark

   Put x = 5000 in 9x − 4y = 5000. Then 9(5000) − 4y = 5000, i.e., 45000 − 4y = 5000, so 4y = 40000 and y = 10000.

   Why? Substituting the known value of x into an original equation gives an equation in y only.

4. 4

   Step 4: Find the incomes

   ½ mark

   9x = 9 × 5000 = ₹45000 and 7x = 7 × 5000 = ₹35000. So the monthly incomes are ₹45000 and ₹35000.

   Why? The incomes are the original multiples of x, so multiply the common value x by 9 and 7.

Marks: 1 + 1 + ½ + ½ = 3 · Split as in the CBSE marking scheme, 2025 · Standard 30/4/1 Q27

1. 1

   Step 1: Check consistency

   1 mark (suggested)

   For the two equations, the coefficients of x and y are (1, 3) and (−2, 3). So a₁/a₂ = 1/(−2) = −0.5 and b₁/b₂ = 3/3 = 1. Since a₁/a₂ ≠ b₁/b₂, the pair is consistent.

   Why? For two lines, unequal ratios a₁/a₂ and b₁/b₂ mean the lines intersect at one point, so the pair has a unique solution.

2. 2

   Step 2: Draw the graph

   1 mark (suggested)

   For x + 3y = 6, two points on the line are (0, 2) and (6, 0). For 3y − 2x = −12, two points on the line are (0, −4) and (3, −2). Plot these points and join them to draw the two lines.

   Why? A straight line is fixed by two points, so two solutions of each equation are enough to draw its graph.

3. 3

   Step 3: Read the solution

   1 mark (suggested)

   The two lines meet at (6, 0). Hence x = 6 and y = 0.

   Why? The point that lies on both lines satisfies both equations, so it is the unique solution of the pair.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Form the equations

   1 mark (suggested)

   Let the two natural numbers be x and y, with x > y. Then x + y = 70 and x − y = 10.

   Why? Each condition of the problem gives one linear equation, and two unknowns need two independent equations.

2. 2

   Step 2: Solve by elimination

   1 mark (suggested)

   Adding the equations: 2x = 80, so x = 40. Then y = 70 − 40 = 30. The natural numbers are 40 and 30.

   Why? Adding the two equations cancels y, reducing the pair to a single equation in x; substituting the value back gives y.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Form the equations

   1 mark (suggested)

   Let the tens digit be x and the units digit be y. Then the number is 10x + y and the reversed number is 10y + x. The product of the digits is 18, and subtracting 27 interchanges the digits, so 10x + y − 27 = 10y + x.

   Why? A two-digit number is written as 10x + y, so the condition about reversing digits becomes an equation in x and y.

2. 2

   Step 2: Simplify the reversal condition

   1 mark (suggested)

   10x + y − 27 = 10y + x gives 10x + y − (10y + x) = 27, i.e., 9x − 9y = 27, so x − y = 3, i.e., x = y + 3.

   Why? Bringing the digit terms together isolates 27 on one side; dividing by 9 gives the simple relation x = y + 3.

3. 3

   Step 3: Find the digits and the number

   1 mark (suggested)

   Substituting x = y + 3 in xy = 18: y(y + 3) = 18, so y² + 3y − 18 = 0, i.e., (y + 6)(y − 3) = 0. Hence y = 3 or y = −6; y = −6 is rejected. So y = 3 and x = y + 3 = 6. The number is 10x + y = 10(6) + 3 = 63.

   Why? Putting x = y + 3 into the product condition gives a quadratic in y; solving it and rejecting the negative root gives the digits, and 10x + y rebuilds the number.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Eliminate y

   1 mark

   Multiply 7x − 2y = 5 by 7 and 8x + 7y = 15 by 2: 49x − 14y = 35 and 16x + 14y = 30. Adding gives 65x = 65, so x = 1.

   Why? Multiplying by 7 and by 2 makes the y-coefficients equal in size and opposite in sign, so adding the equations cancels y and leaves one equation in x.

2. 2

   Step 2: Find y

   ½ mark

   Substitute x = 1 in 7x − 2y = 5: 7(1) − 2y = 5, so 2y = 2 and y = 1.

   Why? Replacing x by its known value reduces an equation in two variables to one in y only; solving it gives the second value.

3. 3

   Step 3: Verify and state the solution

   ½ mark

   7(1) − 2(1) = 5 and 8(1) + 7(1) = 15. Hence x = 1, y = 1.

   Why? A pair of values is the solution only if both original equations become true statements; substituting checks this.

Marks: 1 + ½ + ½ = 2 · Split as in the CBSE marking scheme, 2024 · Standard 30/1/1 Q21

1. 1

   Step 1: Eliminate q

   1 mark (suggested)

   Multiply the first equation by 4 and the second by 3: 8p + 12q = 52 15p − 12q = −6 Adding: 23p = 46, so p = 2.

   Why? Making the coefficients of q equal in size and opposite in sign lets addition cancel q, so the pair reduces to one linear equation in p.

2. 2

   Step 2: Find q

   1 mark (suggested)

   Put p = 2 in 2p + 3q = 13: 4 + 3q = 13, so 3q = 9, hence q = 3. Therefore, p = 2 and q = 3.

   Why? Putting the known value of p into one original equation changes it into a single linear equation for q.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)