Pair of Linear Equations 2-Mark Questions Class 10: CBSE PYQs with Answers

Multiply the first equation by 6 and the second by 3 to clear the denominators.

3x + 4y = −6 … (1)

3x − y = 9 … (2)

Add 4 × (2) to (1): 15x = 30, so x = 2.

Substitute x = 2 in (1): 4y = −12. Hence, y = −3.

x = 2; y = −3

2x + 3y = 12 … (1)

x − 2y = −1 … (2)

Add 3 × (2) to 2 × (1): 7x = 21, so x = 3.

Substitute x = 3 in (1): 3y = 6. Hence, y = 2.

x = 3; y = 2

Adding: 47x + 47y = 47, so x + y = 1 … (1).

Subtracting the first equation from the second: x − y = 1 … (2).

Adding (1) and (2): 2x = 2, so x = 1.

From (1), y = 1 − 1 = 0.

x = 1; y = 0

Multiply the first equation by 2 and the second by 3.

2x + y = 8 … (1)

x + 6y = 15 … (2)

Subtract (2) from 6 × (1): 11x = 33, so x = 3.

Substitute x = 3 in (1): y = 2.

x = 3; y = 2

7x − 2y = 5 … (1)

8x + 7y = 15 … (2)

Add 2 × (2) to 7 × (1): 65x = 65, so x = 1.

Substitute x = 1 in (1): −2y = −2. Hence, y = 1.

Check: 7(1) + (−2)(1) = 5; 8(1) + (7)(1) = 15.

x = 1; y = 1

2p + 3q = 13 … (1)

5p − 4q = −2 … (2)

Add 3 × (2) to 4 × (1): 23p = 46, so p = 2.

Substitute p = 2 in (1): 3q = 9. Hence, q = 3.

p = 2; q = 3

2x + 5y = −4 … (1)

4x − 3y = 5 … (2)

Add 5 × (2) to 3 × (1): 26x = 13, so x = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$.

Substitute x = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ in (1): 5y = −5. Hence, y = −1.

x = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$; y = −1

Multiply both equations by 10.

x + 3y = 10 … (1)

2x − y = −1 … (2)

Add 3 × (2) to (1): 7x = 7, so x = 1.

Substitute x = 1 in (1): 3y = 9. Hence, y = 3.

x = 1; y = 3

3x + 5y = 8 … (1)

5x − 3y = 2 … (2)

Add 5 × (2) to 3 × (1): 34x = 34, so x = 1.

Substitute x = 1 in (1): 5y = 5. Hence, y = 1.

x = 1; y = 1

2x − 3y = 17 … (1)

4x + y = 13 … (2)

Add 3 × (2) to (1): 14x = 56, so x = 4.

Substitute x = 4 in (1): −3y = 9. Hence, y = −3.

x = 4; y = −3

3x + 2y = 11 … (1)

2x + 3y = 4 … (2)

Subtract 2 × (2) from 3 × (1): 5x = 25, so x = 5.

Substitute x = 5 in (1): 2y = −4. Hence, y = −2.

x = 5; y = −2

30x + 44y = 10 … (1)

40x + 55y = 13 … (2)

Subtract 4 × (2) from 5 × (1): −10x = −2, so x = $\frac{\text{(}\text{1}\text{)/(}}{\text{5}\text{)}}$.

Substitute x = $\frac{\text{(}\text{1}\text{)/(}}{\text{5}\text{)}}$ in (1): 44y = 4. Hence, y = $\frac{\text{(}\text{1}\text{)/(}}{\text{11}\text{)}}$.

x = $\frac{\text{(}\text{1}\text{)/(}}{\text{5}\text{)}}$; y = $\frac{\text{(}\text{1}\text{)/(}}{\text{11}\text{)}}$

Adding: 100x + 100y = 300, so x + y = 3 … (1).

Subtracting the first equation from the second: 26x − 26y = 26, so x − y = 1 … (2).

Adding (1) and (2): 2x = 4, so x = 2.

From (1), y = 1.

x = 2; y = 1

Adding: 110x − 110y = 110, so x − y = 1 … (1).

Subtracting the second equation from the first: 36x + 36y = 108, so x + y = 3 … (2).

Adding (1) and (2): 2x = 4, so x = 2.

From (2), y = 1.

x = 2; y = 1

Adding: 100x + 100y = 1000, so x + y = 10 … (1).

Subtracting the first equation from the second: 2x − 2y = 2, so x − y = 1 … (2).

Adding (1) and (2): 2x = 11, so x = $\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$.

From (1), y = 10 − $\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{9}\text{)/(}}{\text{2}\text{)}}$.

x = $\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$; y = $\frac{\text{(}\text{9}\text{)/(}}{\text{2}\text{)}}$

x + 2y = 9 … (1)

−2x + y = 2 … (2)

Subtract 2 × (2) from (1): 5x = 5, so x = 1.

Substitute x = 1 in (1): 2y = 8. Hence, y = 4.

x = 1; y = 4

At (−4, 3), x + y + 1 = −4 + 3 + 1 = 0, so the point lies on the first line.

For the second line, x − y = −4 − 3 = −7 ≠ 1.

The point lies on the first line, but not on both lines

The equations are 2x + y = 320 … (1) and 4x + 2y = 600 … (2).

Multiply (1) by 2: 4x + 2y = 640.

Subtract (2): 0 = 40, a contradiction.

2x + y = 320 and 4x + 2y = 600; the system is inconsistent

√2 x + √3 y = 5 … (1).

√3 x − 2√2 y = −√6 … (2).

Multiply (1) by √3 and (2) by √2: √6 x + 3y = 5√3 and √6 x − 4y = −2√3.

Subtract the second new equation from the first: 7y = 7√3, so y = √3.

In (1), √2 x + 3 = 5, so x = 2/√2 = √2.

x = √2; y = √3

x + y = 6 … (1)

2x − 3y = 4 … (2)

Add (2) to 3 × (1): 5x = 22, so x = $\frac{\text{(}\text{22}\text{)/(}}{\text{5}\text{)}}$.

Substitute x = $\frac{\text{(}\text{22}\text{)/(}}{\text{5}\text{)}}$ in (1): y = $\frac{\text{(}\text{8}\text{)/(}}{\text{5}\text{)}}$.

x = $\frac{\text{(}\text{22}\text{)/(}}{\text{5}\text{)}}$; y = $\frac{\text{(}\text{8}\text{)/(}}{\text{5}\text{)}}$

Multiply the first equation by −2: −10x + 6y = −22.

The second equation has the same left-hand side, but equals 22.

Subtracting gives 0 = 44, a contradiction.

The pair is inconsistent; there is no solution

The ratios of the x- and y-coefficients are $\frac{\text{(}\text{3}\text{)/(}}{\text{6}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ and $\frac{\text{(}\text{2}\text{)/(}}{\text{−4}\text{)}}$ = $\frac{\text{(}\text{−1}\text{)/(}}{\text{2}\text{)}}$.

Since $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ ≠ $\frac{\text{(}\text{−1}\text{)/(}}{\text{2}\text{)}}$, the two lines intersect at one point.

The pair is consistent with a unique solution

x + 2y = 6 … (1)

2x + y = 6 … (2)

Subtract 2 × (2) from (1): −3x = −6, so x = 2.

Substitute x = 2 in (1): 2y = 4. Hence, y = 2.

Use y = px + 2: 2 = 2p + 2, so 2p = 0.

x = 2; y = 2; p = 0

For infinitely many solutions, the two equations must represent the same line.

k = 0 gives y = 0 and x = 1, a unique solution; hence k ≠ 0.

The coefficient and constant ratios must be equal: $\frac{\text{(}\text{k}\text{)/(}}{\text{1}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{k}\text{)}}$ = $\frac{\text{(}\text{k²}\text{)/(}}{\text{1}\text{)}}$.

The first equality gives k² = 1, so k = 1 or −1.

At k = 1, both equations are x + y = 1.

At k = −1, the equations are −x + y = 1 and x − y = 1; adding gives 0 = 2, so reject −1.

k = 1

For infinitely many solutions, both equations must represent the same line.

c = 0 gives a unique solution, so c ≠ 0.

$\frac{\text{(}\text{c}\text{)/(}}{\text{12}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{c}\text{)}}$ = $\frac{\text{(}\text{c − 3}\text{)/(}}{\text{c}\text{)}}$.

From $\frac{\text{(}\text{3}\text{)/(}}{\text{c}\text{)}}$ = $\frac{\text{(}\text{c − 3}\text{)/(}}{\text{c}\text{)}}$, 3 = c − 3, so c = 6.

At c = 6, the equations are 6x + 3y = 3 and 12x + 6y = 6; the second is twice the first.

c = 6

3x + 2y = 65 … (1)

2x + 3y = 60 … (2)

Subtract 2 × (2) from 3 × (1): 5x = 75, so x = 15.

Substitute x = 15 in (1): 2y = 20. Hence, y = 10.

x = 15; y = 10

Multiply the second equation by 3: 6x + 9y + 12 = 0.

Subtract the first equation: (9 − k)y + 3 = 0.

If k ≠ 9, this determines exactly one y and then one x.

At k = 9, the subtraction gives 3 = 0, so there is no solution.

Every real k except 9

Multiply the second original equation by 2 to remove its denominator.

7x − 2y = 3 … (1)

22x − 3y = 16 … (2)

Subtract 2 × (2) from 3 × (1): −23x = −23, so x = 1.

Substitute x = 1 in (1): −2y = −4. Hence, y = 2.

x = 1; y = 2

For infinitely many solutions, both equations must represent the same line.

k = 0 gives a unique solution, so k ≠ 0.

$\frac{\text{(}\text{k}\text{)/(}}{\text{12}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{k}\text{)}}$ = $\frac{\text{(}\text{k − 3}\text{)/(}}{\text{k}\text{)}}$.

From $\frac{\text{(}\text{3}\text{)/(}}{\text{k}\text{)}}$ = $\frac{\text{(}\text{k − 3}\text{)/(}}{\text{k}\text{)}}$, 3 = k − 3, so k = 6.

At k = 6, the equations are 6x + 3y = 3 and 12x + 6y = 6; the second is twice the first.

k = 6

For x = −6, plot (−6, 0) and (−6, 8) and draw the vertical line.

For x + y = 2, plot (0, 2), (2, 0) and (−6, 8), then join them.

The two lines meet at (−6, 8).

x = −6; y = 8

Let the cost of one chair be ₹x and one table be ₹y.

x + y = 2000 and y = 3x.

x + 3x = 2000, so x = 500.

y = 3 × 500 = 1500.

One table costs ₹1,500; one chair costs ₹500

For 2x − 3y = −6, plot (−3, 0), (0, 2) and (3, 4) and draw the line.

For x = 3, plot (3, 0) and (3, 4) and draw the vertical line.

The lines intersect at (3, 4).

x = 3; y = 4

For x + 2y = 10, plot (0, 5), (4, 3) and (10, 0), then draw the line.

For y = 3, plot (0, 3) and (4, 3) and draw the horizontal line.

The lines intersect at (4, 3).

x = 4; y = 3

For x + y = 5, plot (0, 5), (4, 1) and (5, 0) and draw the line.

For x − y = 3, plot (0, −3), (3, 0) and (4, 1) and draw the line.

The lines intersect at (4, 1).

x = 4; y = 1

For x = 3, plot (3, 0) and (3, −4), then draw the vertical line.

For y = −4, plot (0, −4) and (3, −4), then draw the horizontal line.

The lines meet at (3, −4).

x = 3; y = −4

x = 0 is the y-axis.

y = −7 is a horizontal line through (0, −7).

The two lines intersect at (0, −7), so there is exactly one solution.

Consistent with a unique solution: x = 0, y = −7

For x = 5, plot (5, 0) and (5, 7), then draw the vertical line.

For y = 7, plot (0, 7) and (5, 7), then draw the horizontal line.

The lines meet at (5, 7).

x = 5; y = 7

x = 0 is the y-axis.

y = −3 is a horizontal line through (0, −3).

The two lines intersect at (0, −3), so there is exactly one solution.

Consistent with a unique solution: x = 0, y = −3

Multiply the first equation by 6 and the second by 3.

9x − 10y = −12 … (1)

−x + 3y = 3 … (2)

Add 10 × (2) to 3 × (1): 17x = −6, so x = $\frac{\text{(}\text{−6}\text{)/(}}{\text{17}\text{)}}$.

Substitute x = $\frac{\text{(}\text{−6}\text{)/(}}{\text{17}\text{)}}$ in (1): −10y = $\frac{\text{(}\text{−150}\text{)/(}}{\text{17}\text{)}}$. Hence, y = $\frac{\text{(}\text{15}\text{)/(}}{\text{17}\text{)}}$.

x = $\frac{\text{(}\text{−6}\text{)/(}}{\text{17}\text{)}}$; y = $\frac{\text{(}\text{15}\text{)/(}}{\text{17}\text{)}}$

Let the greater number be x and the smaller be y.

x − y = 26 and x = 3y.

Substitute x = 3y: 3y − y = 26, so y = 13.

x = 3 × 13 = 39.

The numbers are 39 and 13

Multiply the first equation by 6 and the second by 3.

3x + 4y = −6 … (1)

3x − y = 15 … (2)

Add 4 × (2) to (1): 15x = 54, so x = $\frac{\text{(}\text{18}\text{)/(}}{\text{5}\text{)}}$.

Substitute x = $\frac{\text{(}\text{18}\text{)/(}}{\text{5}\text{)}}$ in (1): 4y = $\frac{\text{(}\text{−84}\text{)/(}}{\text{5}\text{)}}$. Hence, y = $\frac{\text{(}\text{−21}\text{)/(}}{\text{5}\text{)}}$.

x = $\frac{\text{(}\text{18}\text{)/(}}{\text{5}\text{)}}$; y = $\frac{\text{(}\text{−21}\text{)/(}}{\text{5}\text{)}}$

Let length = x cm and breadth = y cm.

2(x + y) = 36, so x + y = 18 … (1).

x − y = 4 … (2).

Adding (1) and (2): 2x = 22, so x = 11.

From (1), y = 18 − 11 = 7.

Length = 11 cm; breadth = 7 cm

Let the greater number be x and the smaller be y.

x + y = 70 … (1); x − y = 10 … (2).

Adding (1) and (2): 2x = 80, so x = 40.

From (1), y = 70 − 40 = 30.

The natural numbers are 40 and 30

x − 3y = 7 … (1)

3x − 3y = 5 … (2)

Subtract (2) from (1): −2x = 2, so x = −1.

Substitute x = −1 in (1): −3y = 8. Hence, y = $\frac{\text{(}\text{−8}\text{)/(}}{\text{3}\text{)}}$.

x = −1; y = $\frac{\text{(}\text{−8}\text{)/(}}{\text{3}\text{)}}$

Adding: 203x + 203y = 609, so x + y = 3 … (1).

Subtracting the first equation from the second: x − y = 1 … (2).

Adding (1) and (2): 2x = 4, so x = 2.

From (1), y = 1.

x = 2; y = 1

Let the greater angle be x° and the smaller be y°.

x + y = 180 … (1); x − y = 50 … (2).

Adding (1) and (2): 2x = 230, so x = 115.

From (1), y = 180 − 115 = 65.

Equations: x + y = 180, x − y = 50; angles: 115° and 65°

Let the greater number be x and the smaller be y.

x + y = 75 … (1); x − y = 25 … (2).

Adding (1) and (2): 2x = 100, so x = 50.

From (1), y = 25.

The numbers are 50 and 25

Add the equations: 6x = 30, so x = 5.

Substitute in 2x + y = 13: y = 13 − 10 = 3.

x − y = 5 − 3 = 2.

x − y = 2

Let the greater number be x and the smaller be y.

x + y = 105 … (1); x − y = 45 … (2).

Adding (1) and (2): 2x = 150, so x = 75.

From (1), y = 30.

The numbers are 75 and 30