CBSE Maths › Class 10 PYQs › Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables: 2 marks Questions (CBSE Class 10)

30 different 2 marks questions on Pair of Linear Equations in Two Variables from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (68)2 marks (30)3 marks (40)4 marks (9)5 marks (29)

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/1/1

Solve the following system of equations for $x$ and $y$ :
$\frac{x}{2} + \frac{2 y}{3} = - 1$ and $x - \frac{y}{3} = 3$

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Answer:

x = 2

y = - 3

Multiply the first equation by 6: $3 x + 4 y = - 6$ ... (1)
Multiply the second equation by 3: $3 x - y = 9$ ... (2)
(1) $-$ (2): $5 y = - 15$ , so $y = - 3$ .
From (2): $3 x = 9 + y = 6$ , so $x = 2$ .

Q242 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/1/2

Solve for $x$ and $y$ :
$0.1 x + 0.3 y = 1$
$0.2 x - 0.1 y = - 0.1$

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Answer:

x = 1

y = 3

Multiply both equations by 10: $x + 3 y = 10$ ... (1) and $2 x - y = - 1$ ... (2)
From (2): $y = 2 x + 1$ .
Substitute in (1): $x + 6 x + 3 = 10$ , so $7 x = 7$ and $x = 1$ .
$y = 2 (1) + 1 = 3$ .

Q222 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/1/3

Solve for $x$ and $y$ :
$3 x + 5 y = 8$
$5 x - 3 y = 2$

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Answer:

x = 1

y = 1

Multiply the first equation by 3 and the second by 5: $9 x + 15 y = 24$ and $25 x - 15 y = 10$ .
Adding: $34 x = 34$ , so $x = 1$ .
Then $3 + 5 y = 8$ , so $y = 1$ .

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/2/1

Find the value of $c$ for which the following pair of linear equations has infinitely many solutions :
$c x + 3 y = c - 3$
$12 x + cy = c$

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Answer:

c = 6

Write as $c x + 3 y - (c - 3) = 0$ and $12 x + cy - c = 0$ .
For infinitely many solutions: $\frac{c}{12} = \frac{3}{c} = \frac{c - 3}{c}$ .
From $\frac{c}{12} = \frac{3}{c}$ : $c^{2} = 36$ , so $c = 6$ or $c = - 6$ .
From $\frac{3}{c} = \frac{c - 3}{c}$ : $3 = c - 3$ , so $c = 6$ .
Check $c = 6$ : $\frac{6}{12} = \frac{3}{6} = \frac{3}{6} = \frac{1}{2}$ . So $c = 6$ .

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q21 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/2/1

Solve for $x$ and $y$ :
$3 x + 2 y = 65$
$2 x + 3 y = 60$

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Answer:

x = 15

y = 10

Adding: $5 x + 5 y = 125 \Rightarrow x + y = 25$ .
Subtracting: $x - y = 5$ .
Adding these: $2 x = 30 \Rightarrow x = 15$ ; then $y = 10$ .
Check: $45 + 20 = 65$ , $30 + 30 = 60$ .

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/3/1

Solve the following system of equations graphically :
$2 x - 3 y = - 6$ and $x = 3$

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Answer:

x = 3

y = 4

$2 x - 3 y = - 6$ : points $(0, 2)$ , $(- 3, 0)$ , $(3, 4)$ . Plot and join them.
$x = 3$ : a vertical line through $(3, 0)$ , parallel to the y-axis.
The two lines meet at $(3, 4)$ .
Hence $x = 3$ , $y = 4$ .

Q242 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/3/2

Solve the following system of equations graphically :
$x + 2 y = 10$ and $y = 3$

Show answer & solution

Answer:

x = 4

y = 3

$x + 2 y = 10$ : points $(0, 5)$ , $(10, 0)$ , $(4, 3)$ . Plot and join them.
$y = 3$ : a horizontal line through $(0, 3)$ , parallel to the x-axis.
The two lines meet at $(4, 3)$ .
Hence $x = 4$ , $y = 3$ .

Q222 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/3/3

Solve the following system of linear equations graphically :
$x + y = 5$ and $x - y = 3$

Show answer & solution

Answer:

x = 4

y = 1

$x + y = 5$ : points $(0, 5)$ , $(5, 0)$ , $(4, 1)$ . Plot and join them.
$x - y = 3$ : points $(3, 0)$ , $(0, - 3)$ , $(4, 1)$ . Plot and join them.
The two lines meet at $(4, 1)$ .
Hence $x = 4$ , $y = 1$ .

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/4/1

Solve the following system of equations algebraically :
$30 x + 44 y = 10$ ; $40 x + 55 y = 13$

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Answer:

x = \frac{1}{5}

y = \frac{1}{11}

Multiply the first equation by 4: $120 x + 176 y = 40$
Multiply the second equation by 3: $120 x + 165 y = 39$
Subtract: $11 y = 1$ , so $y = \frac{1}{11}$
Substitute: $30 x + 4 = 10$ , so $x = \frac{1}{5}$
Check: $40 \times \frac{1}{5} + 55 \times \frac{1}{11} = 8 + 5 = 13$

Q232 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/4/2

Solve the following system of equations algebraically :
$37 x + 63 y = 137$
$63 x + 37 y = 163$

Show answer & solution

Answer:

x = 2

y = 1

Add: $100 x + 100 y = 300 \Rightarrow x + y = 3$ ... (1)
Subtract the first from the second: $26 x - 26 y = 26 \Rightarrow x - y = 1$ ... (2)
From (1) and (2): $x = 2$ , $y = 1$
Check: $37 (2) + 63 (1) = 137$

Q252 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/4/3

Solve the following system of equations algebraically :
$73 x - 37 y = 109$
$37 x - 73 y = 1$

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Answer:

x = 2

y = 1

Subtract: $36 x + 36 y = 108 \Rightarrow x + y = 3$ ... (1)
Add: $110 x - 110 y = 110 \Rightarrow x - y = 1$ ... (2)
From (1) and (2): $x = 2$ , $y = 1$
Check: $73 (2) - 37 (1) = 109$ , $37 (2) - 73 (1) = 1$

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/5/1

The cost of 2 kg apples and 1 kg of grapes on a day was found to be ₹ 320. The cost of 4 kg apples and 2 kg grapes was found to be ₹ 600. If cost of 1 kg of apples and 1 kg of grapes is ₹ $x$ and ₹ $y$ respectively, represent the given situation algebraically as a system of equations and check whether the system so obtained is consistent or not.

Show answer & solution

Answer:

2 x + y = 320

4 x + 2 y = 600

; the system is inconsistent.

Equations: $2 x + y = 320$ and $4 x + 2 y = 600$
$\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{320}{600} = \frac{8}{15}$
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , so the lines are parallel.
The system has no solution: it is inconsistent.

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q21 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/5/1

Solve for $x$ and $y$ :
$2 x + 3 y = 5$ and
$3 x - 8 y = - 6$

Show answer & solution

Answer:

x = 2

y = 3

Multiply the first equation by $3$ : $6 x + 3 y = 53$ ... (1)
Multiply the second equation by $2$ : $6 x - 4 y = - 23$ ... (2)
(1) $-$ (2): $7 y = 73 \Rightarrow y = 3$
Then $2 x + 3 = 5 \Rightarrow x = \frac{2}{2} = 2$
Check: $3 \cdot 2 - 22 \cdot 3 = - 6$ ✓

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/6/1

Solve the following pair of equations algebraically :
$101 x + 102 y = 304$
$102 x + 101 y = 305$

Show answer & solution

Answer:

x = 2

y = 1

Adding: $203 x + 203 y = 609 \Rightarrow x + y = 3$ ... (1)
Subtracting the first from the second: $x - y = 1$ ... (2)
Adding (1) and (2): $2 x = 4 \Rightarrow x = 2$
From (1): $y = 1$

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q21 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/6/1

In a pair of supplementary angles, the greater angle exceeds the smaller by $5 0^{\circ}$ . Express the given situation as a system of linear equations in two variables and hence obtain the measure of each angle.

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Answer:

x + y = 180

x - y = 50

; angles are

11 5^{\circ}

and

6 5^{\circ}

Let the greater angle be $x^{\circ}$ and the smaller be $y^{\circ}$ .
Supplementary: $x + y = 180$ ... (1)
Greater exceeds smaller by $5 0^{\circ}$ : $x - y = 50$ ... (2)
Adding: $2 x = 230 \Rightarrow x = 115$ ; then $y = 65$
The angles are $11 5^{\circ}$ and $6 5^{\circ}$ .

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q252 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/1/2

Solve the following pair of linear equations for $x$ and $y$ algebraically :
$x + 2 y = 9$ and $y - 2 x = 2$

Show answer & solution

Answer:

x = 1

y = 4

From the first equation, $x = 9 - 2 y$ .
Substitute in the second: $y - 2 (9 - 2 y) = 2$ , so $5 y - 18 = 2$ and $y = 4$ .
Then $x = 9 - 8 = 1$ .

Also asked in: 2024 Basic 430/1/3

Q25 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/1/2

Check whether the point $(- 4, 3)$ lies on both the lines represented by the linear equations $x + y + 1 = 0$ and $x - y = 1$ .

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Answer: No; it lies on

x + y + 1 = 0

but not on

x - y = 1

For $x + y + 1 = 0$ : $- 4 + 3 + 1 = 0$ , so the point lies on this line.
For $x - y = 1$ : $- 4 - 3 = - 7 \neq = 1$ , so it does not lie on this line.
Hence the point does not lie on both lines.

Also asked in: 2024 Basic 430/1/3

Q222 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/2/1

The sum of two natural numbers is 70 and their difference is 10. Find the natural numbers.

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Answer: 40 and 30

Let the numbers be $x$ and $y$ ( $x > y$ ).
$x + y = 70$ and $x - y = 10$
Adding: $2 x = 80$ , so $x = 40$
$y = 70 - 40 = 30$
The numbers are 40 and 30.

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q22 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/2/1

Solve for $x$ and y :
$x - 3 y = 7$
$3 x - 3 y = 5$

Show answer & solution

Answer:

x = - 1

y = - \frac{8}{3}

Subtract the first equation from the second: $(3 x - 3 y) - (x - 3 y) = 5 - 7$
$2 x = - 2$ , so $x = - 1$
Then $- 1 - 3 y = 7$ , so $- 3 y = 8$
$y = - \frac{8}{3}$

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/1/1

Solve the following system of linear equations
$7 x - 2 y = 5$ and $8 x + 7 y = 15$ and verify your answer.

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Answer:

x = 1

y = 1

$7 x - 2 y = 5$ ... (1), $8 x + 7 y = 15$ ... (2)
(1) $\times 7$ : $49 x - 14 y = 35$ ; (2) $\times 2$ : $16 x + 14 y = 30$ .
Adding: $65 x = 65$ , so $x = 1$ .
From (1): $7 - 2 y = 5$ , so $y = 1$ .
Verification: $7 (1) - 2 (1) = 5$ and $8 (1) + 7 (1) = 15$ . Both equations hold.

Q232 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/1/2

Solve the following system of linear equations :
$2 p + 3 q = 13$ and $5 p - 4 q = - 2$

Show answer & solution

Answer:

p = 2

q = 3

$2 p + 3 q = 13$ ... (1), $5 p - 4 q = - 2$ ... (2)
(1) $\times 4$ : $8 p + 12 q = 52$ ; (2) $\times 3$ : $15 p - 12 q = - 6$ .
Adding: $23 p = 46$ , so $p = 2$ .
From (1): $4 + 3 q = 13$ , so $q = 3$ .

Q242 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/1/3

Solve the following system of linear equations algebraically :
$2 x + 5 y = - 4$ ; $4 x - 3 y = 5$

Show answer & solution

Answer:

x = \frac{1}{2}

y = - 1

$2 x + 5 y = - 4$ ... (1), $4 x - 3 y = 5$ ... (2)
(1) $\times 2$ : $4 x + 10 y = - 8$ ... (3)
(3) $-$ (2): $13 y = - 13$ , so $y = - 1$ .
From (1): $2 x - 5 = - 4$ , so $x = \frac{1}{2}$ .

Q222 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/3/1

If $2 x + y = 13$ and $4 x - y = 17$ , find the value of $(x - y)$ .

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Answer:

x - y = 2

Adding the equations: $6 x = 30$ , so $x = 5$ .
Then $y = 13 - 2 (5) = 3$ .
$x - y = 5 - 3 = 2$ .

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q22 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/3/1

Sum of two numbers is 105 and their difference is 45. Find the numbers.

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Answer: 75 and 30

Let the numbers be $x$ and $y$ with $x > y$ .
$x + y = 105$ and $x - y = 45$ .
Adding: $2 x = 150$ , so $x = 75$ ; then $y = 105 - 75 = 30$ .
The numbers are 75 and 30.

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q242 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/1/1

Solve for $x$ and $y$ : $x + y = 6$ , $2 x - 3 y = 4$ .

Show answer & solution

Answer:

x = \frac{22}{5}

y = \frac{8}{5}

From the first equation, $x = 6 - y$ .
Substitute: $2 (6 - y) - 3 y = 4$ , so $12 - 5 y = 4$ .
$y = \frac{8}{5}$
$x = 6 - \frac{8}{5} = \frac{22}{5}$

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q24 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/1/1

Find out whether the following pair of linear equations are consistent or inconsistent :
$5 x - 3 y = 11$ , $- 10 x + 6 y = 22$

Show answer & solution

Answer: Inconsistent

Write as $5 x - 3 y - 11 = 0$ and $- 10 x + 6 y - 22 = 0$ .
$\frac{a _{1}}{a _{2}} = \frac{5}{- 10} = - \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 3}{6} = - \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{- 11}{- 22} = \frac{1}{2}$
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , so the lines are parallel.
The pair has no solution, so it is inconsistent.