Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  NCERT Solutions: Pair of Linear Equations in Two Variables (Exercise 3.2)

NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Exercise 3.2)

Q1: Solve the following pair of linear equations by the substitution method

(i) x + y = 14
x – y = 4

Sol: Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get, x = 14 – y
Now, substitute the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the exact value of x;
∵ x = 14 – y
∴ x = 14 – 5 Or x = 9
Hence, x = 9 and y = 5.

(ii) s – t = 3
(s / 3) + (t / 2) = 6

Sol: Given,
s – t = 3 and (s / 3) + (t / 2) = 6 are the two equations.
From 1st equation, we get, s = 3 + t .......(1)
Now, substitute the value of s in second equation to get,
(3 + t) / 3 + (t / 2) = 6
⇒ (2(3 + t) + 3t ) / 6 = 6
⇒ (6 + 2t + 3t) / 6 = 6
⇒ (6 + 5t) = 36
⇒ 5t = 30
⇒ 
t = 6
Now, substitute the value of t in equation (1)
s = 3 + 6 = 9
Therefore, s = 9 and t = 6.
(iii) 3x – y = 3
9x – 3y = 9

Sol: Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3 + y) / 3
Now, substitute the value of x in the given second equation to get,
9(3 + y) / 3 – 3y = 9
⇒ 9 + 3y - 3y = 9
⇒ 9 = 9
Therefore, y has infinite values and since, x = (3 + y) / 3, so x also has infinite values.

(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

Sol: Given,
0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 are the two equations.
From 1st equation, we get,
x = (1.3 - 0.3y) / 0.2 .......(1)
Now, substitute the value of x in the given second equation to get,
0.4(1.3 - 0.3y) / 0.2 + 0.5y = 2.3
⇒ 2(1.3 – 0.3y) + 0.5y = 2.3
⇒ 2.6 – 0.6y + 0.5y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3
Now, substitute the value of y in equation (1), we get,
x = (1.3 - 0.3(3)) / 0.2 = (1.3 - 0.9) / 0.2 = 0.4 / 0.2 = 2
Therefore, x = 2 and y = 3.

(v) √2 x + √3 y = 0
√3 x - √8 y = 0

 Sol: Given,
√2 x + √3 y = 0 and √3 x – √8 y = 0
are the two equations.
From 1st equation, we get,
x = – (√3 / √2)y ..........(1)
Putting the value of x in the given second equation to get,
√3(-√3 / √2)y – √8y = 0
⇒ (-3 / √2)y - √8 y = 0
⇒ y = 0
Now, substitute the value of y in equation (1), we get,
x = 0
Therefore, x = 0 and y = 0.

(vi) (3x / 2) – (5y / 3) = -2
(x / 3) + (y / 2) = (13 / 6)
Sol: Given,
(3x / 2) - (5y / 3) = -2 and (x / 3) + (y / 2) = 13 / 6 are the two equations.
From 1st equation, we get,
(3 / 2)x = -2 + (5y / 3)
⇒ x = 2(-6 + 5y) / 9 = (-12 + 10y) / 9 …………(1)
Putting the value of x in the given second equation to get,
((-12 + 10y) / 9) / 3 + y / 2 = 13 / 6
⇒ y / 2 = 13 / 6 – ((-12 + 10y) / 27 ) + y / 2 = 13 / 6
NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Exercise 3.2)
Now, substitute the value of y in equation (1), we get,
(3x / 2) – 5(3) / 3 = -2
⇒ (3x / 2) – 5 = -2
⇒ x = 2
Therefore, x = 2 and y = 3.

Q2: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Sol: 
2x + 3y = 11………………(I)
2x – 4y = -24……………(II)
From equation (II), we get
x = (11 - 3y) / 2 ……………….(III)
Substituting the value of x in equation (II), we get
2(11 - 3y) / 2 – 4y = 24
11 – 3y – 4y = -24
-7y = -35
y = 5…………………(IV)
Putting the value of y in equation (III), we get
x = (11 - 3 × 5) / 2 = -4 / 2 = -2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore the value of m is -1.

Q3: 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.
Sol: Let the two numbers be x and y respectively, such that y > x.
According to the question,
y = 3x ……………… (1)
y – x = 26 …………..(2)
Substituting the value of (1) into (2), we get
3x – x = 26
x = 13 ……………. (3)
Substituting (3) in (1), we get y = 39
Hence, the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Sol: Let the larger angle by xº and smaller angle be yº.
We know that the sum of two supplementary pair of angles is always 180º.
According to the question,
x + y = 180º……………. (1)
x – y = 18º ……………..(2)
From (1), we get x = 180º – y …………. (3)
Substituting (3) in (2), we get
180º – y – y =18º
162º = 2y
y = 81º ………….. (4)
Using the value of y in (3), we get
x = 180º – 81º
= 99º
Hence, the angles are 99º and 81º.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.
Solution: Let the cost a bat be x and cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (I)
3x + 5y = 1750 ………………. (II)
From (I), we get
y = (3800 - 7x) / 6………………..(III)
Substituting (III) in (II). we get,
3x + 5(3800 - 7x) / 6 =1750
⇒ 3x + 9500 / 3 – 35x / 6 = 1750
⇒ 3x - 35x / 6 = 1750 – 9500 / 3
⇒ (18x - 35x) / 6 = (5250 – 9500) / 3
⇒ -17x / 6 = -4250 / 3
⇒ -17x = -8500
x = 500 ……………………….. (IV)
Substituting the value of x in (III), we get
y = (3800 - 7 × 500) / 6 = 300 / 6 = 50
Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 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 Rs 105 and for a journey of 15 km, the charge paid is Rs 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?
Solution: Let the fixed charge be Rs x and per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. (1)
x + 15y = 155 …………….. (2)
From (1), we get x = 105 – 10y ………………. (3)
Substituting the value of x in (2), we get
105 – 10y + 15y = 155
5y = 50 y = 10 …………….. (4)
Putting the value of y in (3), we get
x = 105 – 10 × 10 = 5
Hence, fixed charge is Rs 5 and per km charge = Rs 10

(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.
Solution: Let the fraction be x / y.
According to the question,
(x + 2) / (y + 2) = 9 / 11
11x + 22 = 9y + 18
11x – 9y = -4 …………….. (1)
(x + 3) / (y + 3) = 5 / 6
6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get x = (-4 + 9y) / 11 …………….. (3)
Substituting the value of x in (2), we get
6(-4 + 9y) / 11 - 5y = -3
-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y in (3), we get
x = (-4 + 9 × 9 ) / 11 = 7
Hence the fraction is 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?
Solution: Let the age of Jacob and his son be x and y respectively.
According to the question, (x + 5) = 3(y + 5) x – 3y = 10 ………………………(1)
(x – 5) = 7(y – 5) x – 7y = -30 ………………………(2)
From (1), we get x = 3y + 10 ……………………. (3)
Substituting the value of x in (2), we get
3y + 10 – 7y = -30
-4y = -40
y = 10 ………………(4)
Substituting the value of y in (3), we get
x = 3 x 10 + 10 = 40
Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

The document NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Exercise 3.2) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Exercise 3.2)

1. What is the meaning of a pair of linear equations in two variables?
Ans. A pair of linear equations in two variables is a set of two equations that involve two variables and are related by an equality or inequality. The solution to the pair of equations is the values of the variables that satisfy both equations simultaneously.
2. How can we solve a pair of linear equations in two variables?
Ans. There are several methods to solve a pair of linear equations in two variables. One common method is the substitution method, where we solve one equation for one variable and substitute the value into the other equation. Another method is the elimination method, where we manipulate the equations to eliminate one variable and solve for the other. Graphical method and matrix method are also used to solve such equations.
3. What is the importance of solving a pair of linear equations in two variables?
Ans. Solving a pair of linear equations in two variables helps us find the common solution that satisfies both equations. This is useful in various real-life situations, such as finding the number of items that need to be produced and sold to maximize profit, determining the intersection points of two lines, or analyzing the relationships between different variables in a system.
4. Can a pair of linear equations in two variables have no solution?
Ans. Yes, a pair of linear equations in two variables can have no solution. This happens when the lines represented by the equations are parallel and never intersect. In such cases, the system of equations is said to be inconsistent.
5. How can we determine the number of solutions of a pair of linear equations in two variables?
Ans. The number of solutions of a pair of linear equations in two variables can be determined by analyzing the slopes and intercepts of the lines represented by the equations. If the lines are distinct and intersect at a point, there is a unique solution. If the lines are coincident, meaning they are the same line, there are infinitely many solutions. If the lines are parallel, there is no solution.
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