NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Class 10 Maths Exercise 3.1

1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
See Solution(i) Let the number of girls = x and let the number of boys = y
Total number of students = 10
Therefore,
x + y = 10     …(1)
According to the question, the number of girls is 4 more than the number of boys,
so, x = y + 4     …(2)
To represent graphically, three solutions of each equation is needed.
From equation (1), we get x = 10 – y
x | 5 | 4 | 6
y | 5 | 6 | 4
From equation (2), we get x = y + 4
x | 5 | 4 | 3
y | 1 | 0 | -1

(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
See Solution(ii) Let the cost of one pencil = ₹x and the cost of one pen = ₹y
According to the first condition:
5x + 7y = 50        …(1)
According to the second condition:
7x + 5y = 46        …(2)
To represent graphically three solutions of each equation is given below,
From equation (1), we get:
x = (50 – 7y)/5
x | 3 | 10 | -4
y | 5 | 0 | 10
From equation (2), we get:
x = (46 – 5y)/7
x | 7 | 3 | -2
y | 8 | -2 | 12

2. On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x − 4y + 8 = 0
7x + 6y − 9 = 0
See Solution(i) 5x − 4y + 8 = 0
7x + 6y − 9 = 0
Here, a₁/a₂ = 5/7 and b₁/b₂ = −4/6 = −2/3
⇒ a₁/a₂ ≠ b₁/b₂
So, the given pairs of linear equations intersect at a point.

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
See Solution(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Here, a₁/a₂ = 9/18 = 1/2,
b₁/b₂ = 3/6 = 1/2, and
c₁/c₂ = 12/24 = 1/2
⇒ a₁/a₂ = b₁/b₂ = c₁/c₂
So, the following pairs of linear equations are coincident.

(iii) 6x − 3y + 10 = 0
2x − y + 9 = 0
See Solution(iii) 6x − 3y + 10 = 0
2x − y + 9 = 0
Here, a₁/a₂ = 6/2 = 3/1,
b₁/b₂ = −3/−1 = 3/1, and
c₁/c₂ = 10/9
⇒ a₁/a₂ = b₁/b₂ ≠ c₁/c₂
So, the following pairs of linear equations are parallel.

3. On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the following pairs of linear equations are consistent or inconsistent:
(i) 3x + 2y = 5 ; 2x − 3y = 7
See Solution(i) 3x + 2y = 5
2x − 3y = 7
Here, a₁/a₂ = ³⁄₂ and b₁/b₂ = ²⁄₋₃
⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent.

(ii) 2x − 3y = 8 ; 4x − 6y = 9
See Solution(ii) 2x − 3y = 8
4x − 6y = 9
Here, a₁/a₂ = ²⁄₄ = ½ , b₁/b₂ = ₋₃⁄₋₆ = ½ , and c₁/c₂ = ⁸⁄₉
⇒ a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so, pair of linear equations are inconsistent.

(iii) (³⁄₂)x + (⁵⁄₃)y = 7 ; 9x − 10y = 14
See Solution(iii) (³⁄₂)x + (⁵⁄₃)y = 7
9x − 10y = 14
Here, a₁/a₂ = (³⁄₂)⁄⁹ = ¹⁄₆ and b₁/b₂ = (⁵⁄₃)⁄₋₁₀ = ₋¹⁄₆
⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent.

(iv) 5x − 3y = 11 ; −10x + 6y = −22
See Solution(iv) 5x − 3y = 11
−10x + 6y = −22
Here, a₁/a₂ = ⁵⁄₋₁₀ = ₋½ , b₁/b₂ = ₋³⁄₆ = ₋½ , and c₁/c₂ = ¹¹⁄₋₂₂ = ₋½
⇒ a₁/a₂ = b₁/b₂ = c₁/c₂, so, pair of linear equations are consistent.

(v) (⁴⁄₃)x + 2y = 8 ; 2x + 3y = 12
See Solution(v) (⁴⁄₃)x + 2y = 8
2x + 3y = 12
Here, a₁/a₂ = (⁴⁄₃)⁄₂ = ²⁄₃ , b₁/b₂ = ²⁄₃ and c₁/c₂ = ⁸⁄₁₂ = ²⁄₃
⇒ a₁/a₂ = b₁/b₂ = c₁/c₂, so, pair of linear equations are consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
See Solution(i) x + y = 5 …(1)
2x + 2y = 10        …(2)
Here, a₁/a₂ = 1/2, b₁/b₂ = 1/2, and c₁/c₂ = 5/10 = 1/2
⇒ a₁/a₂ = b₁/b₂ = c₁/c₂, so, pairs of linear equations are consistent.
For three solutions of each equation,
From equation (1), we get, x = 5 − y
x | 4 | 3 | 2
y | 1 | 2 | 3
From equation (2), we get, x = (10 − 2y)/2
x | 4 | 3 | 2
y | 1 | 2 | 3

(ii) x − y = 8, 3x − 3y = 16
See Solution(ii) x − y = 8  …(1)
3x − 3y = 16 …(2)
Here, a₁/a₂ = 1/3,
b₁/b₂ = -1/-3 = 1/3, and c₁/c₂ = 8/16 = 1/2
⇒ a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so, pairs of linear equations are inconsistent.

(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
See Solution(iii) 2x + y − 6 = 0  …(1)
4x − 2y − 4 = 0  …(2)
Here, a₁/a₂ = 2/4 = 1/2 and b₁/b₂ = 1/(-2) = -1/2
⇒ a₁/a₂ ≠ b₁/b₂, so, pairs of linear equations are consistent/inconsistent
For three solutions of each equation,
From equation (1), we get, x = (6 − y)/2
x | 0 | 1 | 2
y | 6 | 4 | 2
From equation (2), we get, x = (4 + 2y)/4
x | 1 | 2 | 3
y | 0 | 2 | 4

(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
See Solution(iv) 2x − 2y − 2 = 0  …(1)
4x − 4y − 5 = 0  …(2)
Here, a₁/a₂ = 2/4 = ¹⁄₂,
b₁/b₂ = (−2)/(−4) = ¹⁄₂, and c₁/c₂ = (−2)/(−5) = ²⁄₅
⇒ a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so, pairs of linear equations are inconsistent.

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
See AnswerLet the breadth of garden = x m and let the length of the garden = y m
Half of the perimeter = 36 m
⇒ 1/2 [2(x + y)] = 36
⇒ x + y = 36 … (1)
According to question, length is 4 m more than its width, so
y = x + 4 … (2)
To get three solutions of each equation,
Form the equation (1), we get x = 36 – y
x | 0 | 36 | 16
y | 36 | 0 | 20
From the equation (2), we get y = x + 4
x | 0 | 8 | 12
y | 4 | 8 | 16

6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
See Solution(i) Given line 2x + 3y – 8 = 0 intersect the line x + 3y – 10 = 0, as
a₁/a₂ ≠ b₁/b₂
(ii) Given line 2x + 3y – 8 = 0 is parallel to 4x + 6y – 9 = 0, as
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
(iii) Given line 2x + 3y – 8 = 0 is coincident to 4x + 6y – 16 = 0, as
a₁/a₂ = b₁/b₂ = c₁/c₂

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
See SolutionFrom the equation (1), we get x = y – 1
x | 0 | 1 | 2
y | 1 | 2 | 3
From the equation (2), we get x = (12 – 2y)/3
x | 4 | 2 | 0
y | 0 | 3 | 6
The coordinates of the vertices of the triangle formed by these lines and the x-axis are (-1, 0), (4, 0) and (2, 3).

Chapter 3 Board Questions

Class 10 Maths Exercise 3.2

1. Solve the following pair of linear equations by the substitution method:
(i) x + y = 14; x – y = 4
See Solution(i) x + y = 14        … (1)
x – y = 4          … (2)
From the equation (1), we get
y = 14 – x       … (3)
Putting the value of y in equation (2), we get
x – (14 – x) = 4
⇒ 2x = 18
⇒ x = 9
Putting the value of x in equation (3), we get
y = 14 – 9 = 5
Hence, x = 9 and y = 5.

(ii) s – t = 3; s/3 + t/2 = 6
See Solution(ii) s – t = 3          … (1)
s/3 + t/2 = 6           … (2)
From the equation (1), we get
s = 3 + t                     … (3)
Putting the value of s in equation (2), we get
(3 + t)/3 + t/2 = 6
⇒ (6 + 2t + 3t) / 6 = 6
⇒ 5t + 6 = 36
⇒ t = 6
Putting the value of t in equation (3), we get
s = 3 + 6 = 9
Hence, s = 9 and t = 6.

(iii) 3x – y = 3; 9x – 3y = 9
See Solution(iii) 3x – y = 3 … (1)
9x – 3y = 9 … (2)
From the equation (1), we get
y = 3x – 3 … (3)
Putting the value of y in equation (2), we get
9x – 3(3x – 3) = 9
⇒ 9 = 9, which is true.
Hence, pair of linear equations have infinite many solution.

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
See Solution(iv) 0.2x + 0.3y = 1.3         … (1)
0.4x + 0.5y = 2.3              … (2)
From the equation (1), we get
y = (1.3 – 0.2x) / 0.3        … (3)
Putting the value of y in equation (2), we get
0.4x + 0.5 [( (1.3 – 0.2x) / 0.3 )] = 2.3
⇒ 0.12x + 0.65 – 0.10x = 0.69
⇒ 0.02x = 0.04
⇒ x = 2
Putting the value of x in equation (3), we get
y = (1.3 – 0.2(2)) / 0.3 = 0.9 / 0.3 = 3
Hence, x = 2 and y = 3.

(v) √2x + √3y = 0; √3x – √2y = 0
See Solution(v) √2x + √3y = 0 … (1)
√3x – √2y = 0 … (2)
From the equation (1), we get
y = -√2x / √3 … (3)
Putting the value of y in equation (2), we get
√3x – √2 [(-√2x / √3)] = 0
⇒ 3x + 2x = 0
⇒ x = 0
Putting the value of x in equation (3), we get
y = 0
Hence, x = 0 and y = 0.

(vi) 3x/2 – 5y/3 = -2; x/3 + y/2 = 13/6
See Solution(vi) 3x/2 – 5y/3 = -2        … (1)
x/3 + y/2 = 13/6        . .. (2)
From the equation (1), we get,
5y/3 = 3x/2 + 2 = (3x + 4)/2
y = (3/5) [((3x+4)/2)] = (9x+12)/10          … (3)
Putting the value of y in equation (2), we get
x/3 + (1/2) [((9x + 12)/10)] = 13/6
⇒ (20x + 27x + 36) / 60 = 13/6
⇒ 282x + 216 = 780
⇒ 282x = 564
⇒ x = 2
Putting the value of x in equation (3), we get
y = (9(2) + 12) / 10
= (18 + 12) / 10
= 30 / 10 = 3
Hence, x = 2 and y = 3.

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
See Solution2x + 3y = 11      … (1)
2x – 4y = -24        … (2)
From the equation (1), we get
y = (11 – 2x) / 3   … (3)
Putting the value of y in equation (2), we get
2x – 4 [( (11 – 2x) / 3)] = -24
⇒ 6x – 44 + 8x = -72
⇒ 14x = -28
⇒ x = -2
Putting the value of x in equation (3), we get
y = (11 – 2(-2))/3 = (11 + 4)/3 = 15/3 = 5
Hence, x = -2 and y = 5.
Putting the value of x and y in y = mx + 3, we get
5 = m(-2) + 3
⇒ 5 = -2m + 3
⇒ 2m = -2
⇒ m = -1

3. Form the pair of linear equations for the following problems and find their solution by substitution method:
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
See Solution(i) Let the first number = x
Let the second number = y
According to question,
x = 3y       … (1)
The difference between two number is 26, therefore
x – y = 26    … (2)
Putting the value of x in equation (2), we get
3y – y = 26
=> 2y = 26
=> y = 13
Putting the value of y in equation (1), we get
x = 3(13) = 39
Hence, one number is 13 and the other number is 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
See Solution(ii) Let the larger angle = x
Let the smaller angle = y
According to question,
x = y + 18       … (1)
Both angles are supplementary, therefore
x + y = 180      … (2)
Putting the value of x in equation (2), we get
y + 18 + y = 180
=> 2y + 18 = 180
=> 2y = 162
=> y = 81
Putting the value of y in equation (1), we get
x = 81 + 18 = 99
Hence, one angle is 81° and the other one is 99°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
See Solution(iii) Let the cost of one bat = ₹x
Let the cost of one ball = ₹y
According to first condition,
7x + 6y = 3800
=> x = (3800 – 6y) / 7           … (1)
According to second condition,
3x + 5y = 1750 … (2)
Putting the value of x in equation (2), we get
3 [(3800 – 6y) / 7] + 5y = 1750
(11400 – 18y) / 7 + 5y = 1750
11400 – 18y + 35y = 12250
11400 + 17y = 12250
=> 17y = 850
=> y = 50
Putting the value of y in equation (1), we get
x = [3800 – 6 (50)] / 7
x = (3800 – 300) / 7
x = 3500 / 7
x = 500
Hence, the cost of one bat is ₹500 and the cost of one ball is ₹50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
See Solution(iv) Let the fixed charges = ₹x
Let the charge per kilometre = ₹y
According to first condition, for travelling 10 km,
x + 10y = 105
=> x = 105 – 10y        … (1)
According to second condition, for travelling 15 km,
x + 15y = 155          … (2)
Putting the value of x in equation (2), we get
(105 – 10y) + 15y = 155
=> 105 + 5y = 155
=> 5y = 50
=> y = 10
Putting the value of y in equation (1), we get
x = 105 – 10 (10) = 105 – 100 = 5
So, the fixed charges x = ₹5 and charge per km y = ₹10.
Therefore, the charges for 25 km = x + 25y = ₹[5 + 25(10)] = ₹(5 + 250) = ₹255

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
See Solution(v) Let the numerator = x
Let the denominator = y
Therefore, the fraction = x/y
According to first condition,
(x + 2) / (y + 2) = 9/11 => 11(x + 2) = 9(y + 2) => 11x + 22 = 9y + 18
=> 11x – 9y = -4
=> x = (9y – 4) / 11 … (1)
According to second condition,
(x + 3) / (y + 3) = 5/6
=> 6(x + 3) = 5(y + 3)
=> 6x + 18 = 5y + 15
=> 6x – 5y = -3 … (2)
Putting the value of x in equation (2), we get
6 [(9y – 4) / 11] – 5y = -3
(54y – 24) / 11 – 5y = -3
54y – 24 – 55y = -33
=> -y – 24 = -33
=> -y = -9 => y = 9
Putting the value of y in equation (1), we get
x = [9(9) – 4] / 11 = (81 – 4) / 11 = 77 / 11 = 7
Hence, the fraction = x/y = 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
See Solution(vi) Let the age of Jacob = x years
Let the age of son = y years
After 5 years,
Jacob’s age = x + 5 years
Son’s age = y + 5 years
According to question,
x + 5 = 3(y + 5)
=> x + 5 = 3y + 15
=> x = 3y + 10           … (1)
5 years ago,
Jacob’s age = x – 5 years
Son’s age = y – 5 years
According to question,
x – 5 = 7(y – 5)
=> x – 5 = 7y – 35
=> x – 7y = -30           … (2)
Putting the value of x in equation (2), we get
3y + 10 – 7y = -30
=> -4y + 10 = -30
=> -4y = -40
=> y = 10
Putting the value of y in equation (1), we get
x = 3(10) + 10 = 30 + 10 = 40
Hence, the age of Jacob is 40 years and the age of his son is 10 years.

Chapter 3 MCQs

Class 10 Maths Exercise 3.3

1. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
See Solution(i) x + y = 5 … (1)
2x – 3y = 4 … (2)
Multiply equation (1) by 2 and subtracting from equation (2), we get
– 5y = -6
=> y = 6/5
Putting the value of y in equation (1), we get
x + 6/5 = 5
=> x = 5 – 6/5 = (25 – 6)/5
=> x = 19/5
Hence, x = 19/5 and y = 6/5.

(ii) 3x + 4y = 10 and 2x – 2y = 2
See Solution(ii) 3x + 4y = 10              … (1)
2x – 2y = 2             … (2)
Multiply equation (2) by 2 and adding with equation (1), we get
7x = 14
=> x = 14/7 = 2
Putting the value of x in equation (1), we get
3(2) + 4y = 10
=> 6 + 4y = 10
=> 4y = 4
=> y = 1
Hence, x = 2 and y = 1.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
See Solution(iii) 3x – 5y – 4 = 0 … (1)
9x – 2y = 7 … (2)
Multiply equation (1) by 3 and subtracting from equation (2), we get
13y = -5
=> y = -5/13
Putting the value of y in equation (2), we get
9x – 2(-5/13) = 7
=> 9x + 10/13 = 7
=> 9x = 7 – 10/13 = (91 – 10)/13 = 81/13
=> x = (81/13) / 9 = 81/[13(9) = 9/13
Hence, x = 9/13 and y = -5/13.

(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
See Solution(iv) x/2 + 2y/3 = -1
=> 3x + 4y = -6 … (1)
and x – y/3 = 3
=> 3x – y = 9 … (2)
Subtracting equation (1) from equation (2), we get
5y = -15
=> y = -3
Putting the value of y in equation (1), we get
3x + 4(-3) = -6
=> 3x – 12 = -6
=> 3x = 6
=> x = 2
Hence, x = 2 and y = -3.

2. Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
See Solution(i) Let the numerator = x
Let the denominator = y
Therefore, the fraction = x/y
According to first condition,
(x + 1) / (y – 1) = 1
=> x + 1 = y – 1
=> x – y = -2          … (1)
According to second condition,
x / (y + 1) = 1/2
=> 2x = y + 1
=> 2x – y = 1         … (2)
Multiply equation (1) by 2 and subtracting from equation (2), we get
y = 5
Putting the value of y in equation (1), we get
x – 5 = -2
=> x = -2 + 5
=> x = 3
Hence, the fraction = x/y = 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
See Solution(ii) Let the age of Nuri = x years
Let the age of Sonu = y years
5 years ago,
Age of Nuri = x – 5 years
Age of Sonu = y – 5 years
According to question, x – 5 = 3(y – 5)
=> x – 5 = 3y – 15
=> x – 3y = -10 … (1)
After 10 years,
Age of Nuri = x + 10 years
Age of Sonu = y + 10 years
According to question, x + 10 = 2(y + 10)
=> x + 10 = 2y + 20
=> x – 2y = 10 … (2)
Subtracting equation (2) from equation (1), we get
-y = -20
=> y = 20
Putting the value of y in equation (1), we get
x – 3(20) = -10
=> x – 60 = -10
=> x = 50
Hence, age of Nuri is 50 years age of Sonu is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
See Solution(iii) Let the one’s place = x
Let the ten’s place = y
Therefore, the number = 10y + x
Sum of digits is 9, therefore
x + y = 9 … (1)
Number obtained by reversing the digits = 10x + y
According to question,
9(10y + x) = 2(10x + y)
=> 90y + 9x = 20x + 2y
=> 88y – 11x = 0
=> 11x – 88y = 0
=> x – 8y = 0 … (2)
Subtracting equation (2) from equation (1), we get
9y = 9
=> y = 1
Putting the value of y in equation (1), we get
x + 1 = 9
=> x = 8
Therefore, the number = 10y + x = 10(1) + 8 = 18
Hence, the two digit number is 18.

(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
See Solution(iv) Let the number of notes of ₹50 = x
Let the number of notes of ₹100 = y
Total number of notes is 25, therefore
x + y = 25 … (1)
The total amount of ₹50 and ₹100 is ₹2000, therefore
50x + 100y = 2000
=> x + 2y = 40 … (2)
Subtracting equation (1) from equation (2), we get
y = 15
Putting the value of y in equation (1), we get
x + 15 = 25
=> x = 10
Hence, the number of ₹50 notes is 10 and the number of ₹100 is 15.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
See Solution(v) Let the charge for first three days = ₹x
Let the additional charge for each day = ₹y
Charge for seven days is ₹27, therefore
x + 4y = 27 … (1)
Charge for 5 days is ₹21, therefore
x + 2y = 21 … (2)
Subtracting equation (2) from equation (1), we get
2y = 6
=> y = 3
Putting the value of y in equation (1), we get
x + 4(3) = 27
=> x + 12 = 27
=> x = 15
Hence, the fixed charge is ₹15 and the additional charge for each day is ₹3.

Exercise 3.1 Solutions
Exercise 3.2 Solutions
Exercise 3.3 Solutions

Class 10 Maths Chapter 3 Board Questions
Previous Years Questions from CBSE Board Exams based on Class 10 Maths Chapter 3 Pair of Liner Equations in Two Variables are given here with answer and explanation. Past Years Exams questions are important to prepare for school test and to cross-check your knowledge about the chapter.

1. Solve the following system of equations algebraically: 30x + 44y = 19; 40x + 55y = 13. [CBSE 2025]
See AnswerMultiplying the first equation by 2: 60x + 88y = 38
Multiplying the second equation by 3: 120x + 165y = 39
Subtracting the first from the second: 60x + 77y = 1
From 60x + 88y = 38:
60x = 38 – 88y
x = (38 – 88y)/60
Substituting into 60x + 77y = 1:
60((38 – 88y)/60) + 77y = 1
38 – 88y + 77y = 1
38 – 11y = 1
-11y = 1 – 38 = -37
y = 37/11
Substituting back: x = (38 – 88(37/11))/60
x = (38 – 296)/60
x = -258/60 = -43/10
Therefore, x = -43/10 and y = 37/11

2. The monthly incomes of two persons are in the ratio 9:7 and their monthly expenditures are in the ratio 4:3. If each saved ₹5,000, express the given situation algebraically as a system of linear equations in two variables. Hence, find their respective monthly incomes. [CBSE 2025]
See AnswerLet’s denote:
x = monthly income of the first person
y = monthly income of the second person
x – 5000 = expenditure of first person
y – 5000 = expenditure of second person
Given:
x/y = 9/7 (income ratio)
(x – 5000)/(y – 5000) = 4/3 (expenditure ratio)
From the first equation: x = (9/7)y
From the second equation: (x – 5000)/(y – 5000) = 4/3
3(x – 5000) = 4(y – 5000)
3x – 15000 = 4y – 20000
3x – 4y = -5000
Substituting x = (9/7)y into the equation:
3(9/7)y – 4y = -5000
27y/7 – 4y = -5000
27y/7 – 28y/7 = -5000
-y/7 = -5000
y = 35000
Now find x:
x = (9/7)y = (9/7) × 35000 = 45000
Therefore, the monthly incomes are ₹45,000 and ₹35,000.

3. If 2x + y = 13 and 4x – y = 17, then find the value of (x – y). [CBSE 2024]
See SolutionAdding the two equations: 2x + y = 13 and 4x – y = 17
6x = 30
Therefore, x = 5
Substituting x = 5 into the first equation: 2(5) + y = 13
10 + y = 13
y = 3
Now I can find (x – y): x – y = 5 – 3 = 2
Therefore, (x – y) = 2

4. Solve the following system of linear equations graphically: x – y + 1 = 0 and x + y = 5 [CBSE 2024]
See SolutionRearranging the equations:
x – y = -1 …(1)
x + y = 5 …(2)
For equation (1): x-intercept (when y = 0): x = -1 y-intercept (when x = 0): -y = -1, so y = 1
For equation (2): x-intercept (when y = 0): x = 5 y-intercept (when x = 0): y = 5
When plotted, these lines intersect at (2, 3), which can be verified by substituting into both equations.
Therefore, the solution is x = 2, y = 3.

5. Sum of two numbers is 105 and their difference is 45. Find the numbers. [CBSE 2024]
See SolutionLet the two numbers be a and b.
Given:
a + b = 105 …(1)
and
a – b = 45 …(2)
Adding equations (1) and (2):
2a = 150
a = 75
Substituting a = 75 in equation (1):
75 + b = 105
b = 30
Therefore, the two numbers are 75 and 30.

6. Two peoples are 16 km apart on a straight road. They start walking at the same time. If they walk towards each other with different speeds, they will meet in 2 h. Had they walked in the same direction with same speeds as before, they would have met in 8 h. Find their walking speeds. [CBSE 2024]
See SolutionLet their speeds be u km/h and v km/h.
When they walk towards each other:
(u + v) × 2 = 16
u + v = 8 …(1)
When they walk in the same direction:
|u – v| × 8 = 16
|u – v| = 2
Since we don’t know which speed is greater, let’s assume u > v:
u – v = 2 …(2)
Solving equations (1) and (2):
u + v = 8
u – v = 2
2u = 10
u = 5
Substituting back:
5 + v = 8
v = 3
Therefore, their walking speeds are 5 km/h and 3 km/h.

7. A fraction becomes 1/3 when 1 is subtracted from the numerator. It becomes 1/4 when 8 is added to the denominator. Find the fraction. [CBSE 2023]
See SolutionLet the fraction be x/y.
From the first condition:
(x – 1)/y = 1/3
3(x – 1) = y
3x – 3 = y …(1)
From the second condition:
x/(y + 8) = 1/4
4x = y + 8 …(2)
Substituting equation (1) into equation (2):
4x = 3x – 3 + 8
4x = 3x + 5
x = 5
Using equation (1): y = 3x – 3 = 3(5) – 3 = 15 – 3 = 12
Therefore, the fraction is 5/12.

8. The age of the father is twice the sum of the ages of his two children. After 20 yr, his age will be equal to the sum of the ages of his children. Find the present age of the father. [CBSE 2023 Compt]
See SolutionLet the present ages of the two children be a and b years, and the father’s present age be F years.
Given:
F = 2(a + b) …(1)
F + 20 = (a + 20) + (b + 20) …(2)
Simplifying equation (2):
F + 20 = a + b + 40
F = a + b + 20 …(3)
From equations (1) and (3): 2(a + b) = a + b + 20
a + b = 20
Substituting back into equation (1): F = 2(20) = 40
Therefore, the present age of the father is 40 years.

9. Solve the following pair of equations graphically: x + 3y = 6, 2x – 3y = 12. Also, find the area of the triangle formed by these lines representing the equations with Y-axis. [CBSE 2023 Compt]
See SolutionFor x + 3y = 6: x-intercept (y = 0): x = 6 y-intercept (x = 0): y = 2
For 2x – 3y = 12: x-intercept (y = 0): x = 6 y-intercept (x = 0): y = -4
The lines intersect at (6, 0), which can be verified by substitution.
The triangle formed with the Y-axis has vertices at (0, 2), (0, -4) and (6, 0).
The area of this triangle
= (1/2) × base × height
= (1/2) × 6 × (2 – (-4))
= (1/2) × 6 × 6
= 18 square units.

10. In a ΔABC, ∠A = x°, ∠B = (3x – 2)° and ∠C = y°. Also, ∠C – ∠B = 9°. Determine the three angles of the triangle. [CBSE 2023 Compt]
See SolutionGiven: ∠A = x° ∠B = (3x – 2)° ∠C = y° ∠C – ∠B = 9°
Since the angles in a triangle sum to 180°: x + (3x – 2) + y = 180 …(1)
From ∠C – ∠B = 9°:
y – (3x – 2) = 9
y = 3x – 2 + 9
y = 3x + 7 …(2)
Substituting equation (2) into equation (1):
x + (3x – 2) + (3x + 7) = 180
x + 3x – 2 + 3x + 7 = 180
7x + 5 = 180
7x = 175
x = 25
Now we can find the other angles:
∠B = 3x – 2 = 3(25) – 2 = 75 – 2 = 73°
∠C = y = 3x + 7 = 3(25) + 7 = 75 + 7 = 82°
To verify: ∠A + ∠B + ∠C = 25° + 73° + 82° = 180°
Also, ∠C – ∠B = 82° – 73° = 9°
Therefore, the three angles of the triangle are 25°, 73° and 82°.

11. Solve the equations x + 2y = 6 and 2x – 5y = 12 graphically. [CBSE 2020]
See SolutionTo solve graphically, we’ll find the intercepts for each equation and plot them.
For equation x + 2y = 6: x-intercept (when y = 0): x = 6 y-intercept (when x = 0): 2y = 6, so y = 3
For equation 2x – 5y = 12: x-intercept (when y = 0): 2x = 12, so x = 6 y-intercept (when x = 0): -5y = 12, so y = -2.4
Plotting these points and drawing the lines, they intersect at (6, 0).
Therefore, the solution is x = 6, y = 0, which can be verified by substituting into both equations.

12. If 2x + y = 23 and 4x – y = 19, then find the value of (5y – 2x) and (y/x – 2). [CBSE 2020]
See SolutionSolving the system of equations:
2x + y = 23 …(1) and
4x – y = 19 …(2)
Adding equations (1) and (2):
6x = 42
x = 7
Substituting x = 7 into equation (1):
2(7) + y = 23
14 + y = 23
y = 9
Now I can calculate: (5y – 2x) = 5(9) – 2(7) = 45 – 14 = 31
(y/x – 2) = (9/7) – 2 = (9/7) – (14/7) = -5/7
Therefore, (5y – 2x) = 31 and (y/x – 2) = -5/7.

13. A man can row a boat downstream 20 km in 2 h and upstream 4 km in 2 h. Find his speed of rowing in still water. Also, find the speed of the stream. [CBSE 2020]
See SolutionLet the man’s rowing speed in still water be u km/h and the stream’s speed be v km/h.
For downstream:
(u + v) × 2 = 20
u + v = 10 …(1)
For upstream:
(u – v) × 2 = 4
u – v = 2 …(2)
Solving equations (1) and (2):
u + v = 10
u – v = 2
Adding (1) and (2)
2u = 12
u = 6
Substituting back:
6 + v = 10
v = 4
Therefore, the man’s rowing speed in still water is 6 km/h, and the stream’s speed is 4 km/h.

14. It can take 12 h to fill a swimming pool using two pipes. If the pipe of largest diameter is used for four hour and the pipe of smaller diameter for 9 h, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately? [CBSE 2020]
See SolutionLet’s say the larger pipe can fill the pool in x hours and the smaller pipe can fill it in y hours.
In 1 hour, the larger pipe fills 1/x of the pool and the smaller pipe fills 1/y of the pool.
From the first condition (both pipes fill the pool in 12 hours):
(1/x + 1/y) × 12 = 1 1/x + 1/y = 1/12 …(1)
From the second condition (large pipe for 4 hours and small pipe for 9 hours fill half the pool): (1/x × 4) + (1/y × 9) = 1/2 4/x + 9/y = 1/2 …(2)
Multiplying equation (1) by 12: 12/x + 12/y = 1 …(3)
From equation (2) × 2: 8/x + 18/y = 1 …(4)
Subtracting equation (4) from equation (3):
4/x – 6/y = 0
4/x = 6/y
2/x = 3/y
2y = 3x
y = 3x/2 …(5)
Substituting equation (5) into equation (1):
1/x + 1/(3x/2) = 1/12
1/x + 2/(3x) = 1/12
(3 + 2)/3x = 1/12
5/(3x) = 1/12
5 = 3x/12
x = 20
Using equation (5):
y = 3x/2 = 3(20)/2 = 30
Therefore, it would take the larger pipe 20 hours and the smaller pipe 30 hours to fill the pool separately.

15. Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are given by 2y – x = 8, 5y – x = 14 and y – 2x = 1. [CBSE 2020]
See SolutionTo find the vertices of the triangle, I need to find the points of intersection of the three lines:
Line 1: 2y – x = 8 → x = 2y – 8
Line 2: 5y – x = 14 → x = 5y – 14
Line 3: y – 2x = 1 → y = 2x + 1
For the first vertex (intersection of lines 1 and 2): 2y – 8 = 5y – 14 -3y = -6 y = 2 Substituting y = 2 into line 1: x = 2(2) – 8 = 4 – 8 = -4 Vertex 1: (-4, 2)
For the second vertex (intersection of lines 2 and 3): 5y – 14 = x 2x + 1 = y Substituting: 5(2x + 1) – 14 = x 10x + 5 – 14 = x 10x – x = 14 – 5 9x = 9 x = 1 y = 2(1) + 1 = 3 Vertex 2: (1, 3)
For the third vertex (intersection of lines 1 and 3): 2y – 8 = x y = 2x + 1 Substituting: 2(2x + 1) – 8 = x 4x + 2 – 8 = x 4x – x = 8 – 2 3x = 6 x = 2 y = 2(2) + 1 = 5 Vertex 3: (2, 5)
Therefore, the coordinates of the triangle’s vertices are (-4, 2), (1, 3) and (2, 5).

16. A father’s age is three times the sum of the ages of his two children. After 5 yr his age will be two times the sum of their ages. Find the present age of the father. [CBSE 2019]
See SolutionLet the present ages of the two children be x and y years. Then the father’s present age = 3(x + y) years.
After 5 years:
The children’s ages will be (x + 5) and (y + 5) years
The father’s age will be 3(x + y) + 5 years
According to the second condition:
3(x + y) + 5 = 2[(x + 5) + (y + 5)]
3(x + y) + 5 = 2(x + y + 10)
3(x + y) + 5 = 2(x + y) + 20
3(x + y) – 2(x + y) = 20 – 5
(x + y) = 15
This means the sum of the children’s ages is 15 years. Therefore, the father’s present age = 3(15) = 45 years.

17. A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/5 when 1 is subtracted from the denominator. Find the fraction. [CBSE 2019]
See SolutionLet the fraction be x/y where x is the numerator and y is the denominator.
According to the first condition:
(x – 2)/y = 1/3
x – 2 = y/3
3x – 6 = y
According to the second condition:
x/(y – 1) = 1/5
5x = y – 1
From the second equation: y = 5x + 1
Substituting this into the first equation:
3x – 6 = 5x + 1
-2x = 7
x = -7/2
Since a fraction should have positive values for a meaningful answer, this solution isn’t appropriate. Let’s double-check our approach.
Actually, we can solve this differently. From the first condition:
(x – 2)/y = 1/3
3(x – 2) = y
3x – 6 = y
From the second condition:
x/(y – 1) = 1/5
5x = y – 1
y = 5x + 1
Now we have two equations for y:
3x – 6 = 5x + 1
-2x = 7
x = -7/2
Since x = -7/2, then y = 3(-7/2) – 6 = -21/2 – 6 = -21/2 – 12/2 = -33/2
The fraction x/y = (-7/2)/(-33/2) = 7/33
Therefore, the required fraction is 7/33.

18. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution. [CBSE 2019]
See SolutionFor a system of linear equations to have a unique solution, the ratio of coefficients must not be equal.
From the given equations:
x + 2y = 5 …(1)
3x + ky + 15 = 0 …(2)
Rearranging equation (2): 3x + ky = -15
For a unique solution, the ratio of coefficients must be unequal: 1/3 ≠ 2/k
This means k ≠ 6
Therefore, the value of k for which the system has a unique solution is any value except 6.

19. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present? [CBSE 2019]
See SolutionLet’s say Sumit’s present age is S years and his son’s present age is P years.
Given: S = 3P …(1)
S + 5 = 2.5(P + 5) …(2)
Substituting equation (1) into equation (2):
3P + 5 = 2.5(P + 5)
3P + 5 = 2.5P + 12.5
3P – 2.5P = 12.5 – 5
0.5P = 7.5
P = 15
Using equation (1): S = 3P = 3 × 15 = 45
Therefore, Sumit’s present age is 45 years.

20. For what value of k, will the following pair of equations have infinitely many solutions: 2x + 3y = 7 and (k + 2)x – 3(1 – k)y = 5k + 1 [CBSE 2019]
See SolutionFor a system of linear equations to have infinitely many solutions, the ratios of coefficients and constants must be equal.
Given equations: 2x + 3y = 7 …(1)
(k + 2)x – 3(1 – k)y = 5k + 1 …(2)
Simplifying equation (2): (k + 2)x – 3y + 3ky = 5k + 1
For infinitely many solutions: 2/(k + 2) = 3/(-3 + 3k) = 7/(5k + 1)
Looking at the first equality: 2/(k + 2) = 3/(-3 + 3k)
2(-3 + 3k) = 3(k + 2)
-6 + 6k = 3k + 6
3k = 12
k = 4
Let’s verify with the second equality: 3/(-3 + 3k) = 7/(5k + 1) With k = 4: 3/(-3 + 12) = 7/(20 + 1)
= 3/9 = 7/21 = 1/3
Therefore, k = 4 will give infinitely many solutions.

21. Find c, if the system of equations cx + 3y + (3 – c) = 0, 12x + cy – c = 0 has infinitely many solutions. [CBSE 2019]
See SolutionFor infinitely many solutions, the equations must be proportional.
Let’s rearrange the equations:
cx + 3y = c – 3 …(1)
12x + cy = c …(2)
For these to be proportional: c/12 = 3/c = (c-3)/c
From c/12 = 3/c:
c² = 36
c = ±6
From 3/c = (c-3)/c:
3 = c – 3
c = 6
Since both conditions must be satisfied, c = 6.
Therefore, c = 6 will give infinitely many solutions.

22. Find the relation between p and q, if x = 3 and y = 1 is the solution of the pair of equations x – 4y + p = 0 and 2x + y – q = 0 [CBSE 2019]
See SolutionSince (3, 1) is a solution to both equations, substituting these values:
From x – 4y + p = 0:
3 – 4(1) + p = 0
3 – 4 + p = 0
p = 1
From 2x + y – q = 0:
2(3) + 1 – q = 0
6 + 1 – q = 0
q = 7
Therefore, the relation between p and q is: p = 1 and q = 7.

Class 10 Mathematics NCERT Chapter 3 exercise solutions provide step-by-step explanations, ensuring that students grasp the concepts of substitution, elimination and cross-multiplication methods. With detailed NCERT 10th Math Chapter 3 solutions PDF, learners can easily access practice questions and solved examples for better preparation. NCERT Class 10 Maths Chapter 3 Exercises textbook solutions cover all exercises, offering clarity on complex problems. For those aiming to improve, referring to Class 10 NCERT Maths Chapter 3 important questions and Class 10 Math 3rd chapter solved problems can boost confidence and improve exam performance.

To perform well in mathematics, students can download the NCERT Class 10 Maths Chapter 3 solutions PDF and practice regularly. These materials not only provide Class 10 Math Chapter 3 question answers but also explain concepts in an easy-to-follow manner. NCERT Book Class 10 Maths Chapter 3 detailed solutions and Class 10 Mathematics Chapter 3 Exercises solution manual serve as perfect companions for exam preparation, ensuring a thorough understanding of solving linear equations.
Students can explore 10th NCERT Maths Chapter 3 practice questions and NCERT Class 10 Math Book Chapter 3 solved exercises to reinforce their learning. For additional guidance, the Class 10 Maths Chapter 3 solution guide and free-to-download Class 10 Maths Chapter 3 solution book are invaluable resources. Accessing these CBSE Class 10 Mathematics Chapter 3 solutions online makes preparation hassle-free and efficient for every aspiring student.

Key Highlights of Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Class 10 Maths Chapter 3 focuses on solving pairs of linear equations using substitution, elimination and cross-multiplication methods. Key points include graphical solutions, consistency conditions and practical problem-solving applications. Understanding these methods, practicing solved examples and mastering important questions are crucial for scoring well in board exams.

According to new syllabus the class 10th math has only three exercise for board exams. Exercise 3.1 is deleted now. The exercise 3.2 become ex. 3.1, Exercise 3.3 become ex. 3.2 and exercise 3.4 become ex. 3.3 now. The split up syllabus for 10th mathematics chapter 3 is as follows:
Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency.
Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution, by elimination. Simple situational problems.

• Now, Number of exercise: 3
• Number of Period needed: 15
• Weightage of the Chapter: 7-8

UP Board students of High School are now using NCERT Textbooks for most of the subjects. Class 10 Mathematics Books for UP Board is same as Books for Class 10 Math in CBSE Board. So, the students of Uttar Pradesh Board can download, UP Board Solutions for class 10 Math Chapter 3 from this page of Tiwari Academy. Solutions are available in Hindi and English Medium. Graphs are given for all the questions if required. Visit to Discussion Forum to ask your questions. You can reply the questions already asked by other users.

In NCERT Solutions Class 10, all exercises are solved in both English as well as in Hindi medium in order to help all type of students for new academic session. In Maths 10, Exercise 3.1, 3.2, and 3.3 solutions, if there is any inconvenient to understand, please inform us, we will try to solve it. All NCERT Solutions are made for the CBSE exam for March, based on latest CBSE Syllabus 2025-26.

CBSE has reduced the syllabus of all subjects in all the classes. The CBSE Syllabus for Class 10 Maths is reduced to 65 percent now. The changes in 10th Maths chapter 3: Linear equations in two variables are given below.

The new CBSE Syllabus for 2025-26 Class 10 Maths Chapter 3
Pair of linear equations in two variables and graphical method of their solution, consistency /inconsistency.
Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution, by elimination. Simple situational problems. Simple problems on equations reducible to linear equations.
Deleted Section from previous Syllabus
Solution of a pair of linear equations in two variables by Cross-multiplication.

1. Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: 3x + y = 7 & 6x + 2y = 8. [CBSE 2016]
2. Find the value of k for which the system of equations 3𝑥 − 4𝑦 = 7; 𝑘𝑥 + 3𝑦 − 5 = 0 has no solution. [CBSE 2014]
3. A father is three times as old as his son. After five years, his age will be two and a half times as old as his son. Represent this situation algebraically only. [CBSE 2013]
4. For which value of p does the pair of equations given below have a unique solutions? 4x + py + 8 = 0; 2x + 2y + 2 = 0. [CBSE 2010, 2011, 2013]
5. For what value of k, the following system of linear equations has no solutions? 3x + y = 1; (2k – 1)x + (k – 1)y =2k + 1. [CBSE 2010, 2011, 2012]

1. Solve for x and y: 11/x – 1/y = 10 & 9/x – 4/y = 5. [CBSE 2016]
2. Solve using cross multiplication method: 5x + 4y – 4 = 0 & x – 12y – 20 = 0. [CBSE 2016]
3. A man has certain notes of denomination ₹ 20 and ₹ 5 which amount to ₹380. If the number of notes of each kind are interchanged, they amount to ₹60 less than before. Find the number of notes of each denomination. [CBSE 2015]
4. Find the value of ‘k’ for which the following system of equations represents a pair of coincident lines: 𝑥 + 2𝑦 = 3; (𝑘 − 1)𝑥 + (𝑘 + 1)𝑦 = 𝑘 + 3. [CBSE 2014]
5. Check graphically, whether the pair of equations x + 3y = 6 & 2x – 3y = 12 is consistent. If so, then solve them graphically. [CBSE 2013]
6. The path of a train A is given by the equation x + 2y – 4 = 0 and path of another train B is given by the equation 2x + 4y – 12 = 0. Represent this situation graphically and find whether the two trains meet each other at some place. [CBSE 2013]
7. Form a pair of linear equations in two variables from the data given and solve it graphically: Tina went to a book shop to get some story books and textbooks. When her friends asked her how many of each she had bought, she answered – ‘The number of textbooks is two more than twice the number of story books bought. Also, the number of textbooks is four less than four times the number of story books bought. Help her friends to find the number of textbooks and story books she had bought. [CBSE 2013]
8. Determine graphically, the coordinates of the vertices of a triangle whose sides are graphs of the equations 2x – 3y + 6 = 0, 2x + 3y – 18 = 0 and y – 2 = 0. Also, find the area of this triangle. [CBSE 2010, 2011]

1. For Uttarakhand flood victims’ two sections A and B of class X contributed ₹ 1500. If the contribution of X A was ₹ 100 less than that of X B, find graphically the amounts contributed by both the sections. [CBSE 2016]
2. Three lines 3x + 5y = 15, 6x – 5y = 30 and x = 0 are enclosing a beautiful triangular park. Find the points of intersection of the lines graphically and the area of the park if all measurements are in km. [CBSE 2016]
3. Some people collected money to be donated in two Old Age Homes. A part of the donation is fixed and remaining depends on the number of old people in the home. If they donated ₹ 14500 in the home having 60 people and ₹ 19500 with 85 people, find the fixed part of donation and the amount donated for each people. What is the inspiration behind this? [CBSE 2016]
4. While teaching about the Indian National flag, teacher asked the students that how many lines are there in Blue colour wheel? One student replies that it is 8 times the number of colours in the flag. While other says that the sum of the number colours in the flag and number of lines in the wheel of the flag is 27. Convert the statements given by the students into linear equation of two variables. Find the number of lines in the wheel. [CBSE 2015]
5. Determine the value of k for which the following system of linear equations has infinite number of solutions: (k – 3)x + 3y = k & kx + ky = 12. [CBSE 2015]
6. Draw the graph of the following pair of linear equation: x + 3y = 6 & 2x – 3y = 12. Find the ratio of the areas of the two triangles formed by first line, x = 0, y = 0 and second line, x = 0, y = 0. [CBSE 2015]
7. Places A and B are 200km apart on a high way. One car starts from A and another from B at the same time. If the cars travel in the same directions at different speeds, they meet in 10 hours. Find the speeds of the two cars. [CBSE 2014]
8. Show graphically that the system of equations𝑥 + 2𝑦 = 4 and 7𝑥 + 4𝑦 = 18 is consistent with a unique solution (2, 1). [CBSE 2014]

• 1. Solve for x and y: 99x + 101y = 1499; 101x + 99y = 1501. [CBSE 2010, 2011, 2012, 2013, 2014]
• 2. The age of father is equal to sum of ages of his 4 children. After 30 years, sum of the ages of the children will be twice the age of the father. Find the age of the father. [CBSE 2013]
• 3. A person can row a boat 8 km upstream and 24 km downstream in 4 hours. He can row 12 km downstream and 12 km upstream in 4 hours. Find the speed of rowing in still water and the speed of the current. [CBSE 2013]
• 4. Solve for x and y: 37x + 43y = 123; 43x + 37y = 117. [CBSE 2010, 2011, 2012]
• 5. Draw the graph of the equations: x – y + 1 = 0 & 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x – axis and shade the triangular region. Also calculate the area of the triangle so formed. [CBSE 2011]
• 6. The sum of a 2 digit number and number obtained by reversing the order of digits is 99. If the digits of the number differ by 3, find the number. [CBSE 2011]
• 7. Check graphically whether the pair of linear equations 4x – y – 8 = 0 and 2x – 3y + 6 = 0 is consistent. Also determine the vertices of the triangle form by these lines and x – axis. [CBSE 2006, 2011]
• 8. The sum of the digits of a two digit number is 9. Nine times this number is twice the number obtained by reversing the digits. Find the number. [CBSE 2010]
• 9. A leading library has a fixed charge for the first three days and an additional charge for each day thereafter. Sarita paid ₹ 27 for a book kept for seven days. While Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. [CBSE 2010]
• 10. Solve the following system of linear equations by elimination method: 6(ax + by) = 2a + 2b and 6(bx – ay) = 3b – 2a. [CBSE 2006, 2004]

History about Linear Equations with two variables. Around 4000 years ago, Babylon knew how to solve a simple linear equation with two variables. Around 200 BC, the Chinese published that “Nine Chapters of the Mathematical Art”, they displayed the ability to solve a system of equations in three variables (Perotti).
Evidence from about 300 BC indicates that the Egyptians also knew how to solve problems involving a system of two equations in two unknown quantities, including quadratic equations.
Euler brought to light the idea that a system of equations doesn’t necessarily have to have a solution (Perotti). He recognized the need for conditions to be placed upon unknown variables in order to find a solution.
With the turn into the 19th century Gauss introduced a procedure to be used for solving a system of linear equations. Cayley, Euler, Sylvester and others changed linear systems into the use of matrices to represent them. Gauss brought his theory to solve systems of equations proving to be the most effective basis for solving unknowns.