NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

NCERT Solutions Class 10 Maths
Chapter – 3 (Pair of Linear Equations in Two Variables)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 3 Pair of Linear Equations in Two Variables Exercise 3.5 has been provided here to help the students in solving the questions from this exercise.

Chapter : 3 Pair of Linear Equations in Two Variables

Exercise – 3.5

1. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using the cross-multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0

(ii) 2x + y = 5
3x + 2y = 8

(iii) 3x – 5y = 20
6x – 10y = 40

(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0

Solutions –

(i) x – 3y – 3 = 0
3x – 9y – 2 = 0
= $\frac{1}{3}$ , $\frac{b_1}{b_2}$ = $\frac{-3}{-9}$ = $\frac{1}{3}$ , $\frac{c_1}{c_2}$ = $\frac{-3}{-2}$ = $\frac{3}{2}$
= $\frac{b_1}{b_2}$ ≠ $\frac{c_1}{c_2}$
Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5
3x + 2y = 8
= $\frac{2}{3}$ , $\frac{b_1}{b_2}$ = $\frac{1}{2}$ , $\frac{c_1}{c_2}$ = $\frac{-5}{-8}$
≠ $\frac{b_1}{b_2}$
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication method,
$\frac{x}{b_1c_2-c_1b_2} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-8-(-10))}=\frac{y}{-15-(-16)}=\frac{1}{4-3}$

$\frac{x}{2} = \frac{y}{1}$ = 1
∴ x = 2 and y =1

(iii) 3x – 5y = 20
6x – 10y = 40
= $\frac{3}{6}$ = $\frac{1}{2}$ , $\frac{b_1}{b_2}$ = $\frac{-5}{-10}$ = $\frac{1}{2}$ , $\frac{c_1}{c_2}$ = $\frac{20}{40}$ = $\frac{1}{2}$

= $\frac{b_1}{b_2}$ = $\frac{c_1}{c_2}$

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0

= $\frac{1}{3}$ , $\frac{b_1}{b_2}$ = $\frac{-3}{-3}$ = 1, $\frac{c_1}{c_2}$ = $\frac{-7}{-15}$

≠ $\frac{b_1}{b_2}$
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication method,
$\frac{x}{b_1c_2-c_1b_2} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{45-21} = \frac{y}{-21-(-15)} = \frac{1}{-3-(-9)}$

$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$

$\frac{x}{24} = \frac{1}{6}$ and $\frac{y}{-6} = \frac{1}{6}$

∴ x = 4 and y = 1

2. (i) For which values of a and b do the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1

Solution –

(i) 2x + 3y – 7 =0
(a – b)x + (a + b)y – (3a + b -2) = 0

= $\frac{2}{a-b}$ , $\frac{b_1}{b_2}$ = $\frac{3}{a+b}$ , $\frac{c_1}{c_2}$ = $\frac{-7}{-(3a + b -2)}$ = $\frac{7}{(3a + b -2)}$
For infinitely many solutions,
= $\frac{b_1}{b_2}$ = $\frac{c_1}{c_2}$

Thus,
$\frac{2}{a-b}$ = $\frac{7}{(3a + b -2)}$
6a + 2b – 4 = 7a – 7b
a – 9b = -4 ………………………………. (i)

$\frac{2}{a-b}$ = $\frac{3}{a+b}$
2a + 2b = 3a – 3b
a – 5b = 0 ……………………………….…. (ii)
Subtracting (i) from (ii), we get
(a – 9b) – (a – 5b) = – 4
– 4b = – 4
b =1
Substituting this eq. in (ii), we get
a – 5 × 1 = 0
a = 5
Thus, at a = 5 and b = 1, the given equations will have infinite solutions.

(ii) 3x + y -1 = 0
(2k – 1)x + (k – 1)y – 2k – 1 = 0
= $\frac{3}{2k - 1}$ , $\frac{b_1}{b_2}$ = $\frac{1}{k-1}$ , $\frac{c_1}{c_2}$ = $\frac{-1}{-2k-1}$ = $\frac{1}{2k + 1}$
For no solutions,
= $\frac{b_1}{b_2}$ ≠ $\frac{c_1}{c_2}$

$\frac{3}{2k - 1}$ = $\frac{1}{k-1}$ ≠ $\frac{1}{2k + 1}$

$\frac{3}{2k - 1}$ = $\frac{1}{k-1}$

3k – 3 = 2k -1
k =2
Therefore, for k = 2, the given pair of linear equations will have no solution.

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods.
8x + 5y = 9
3x + 2y = 4
Solution –
8x + 5y = 9 ………………….. (i)
3x + 2y = 4 ……………….…. (ii)
From equation (ii), we get

x = $\frac{4-2y}{3}$ ……………………. (iii)
Using this value in equation (i), we get
$\frac{8(4-2y)}{3}$ + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5 ………………………………. (iv)
Using this value in equation (ii), we get
3x + 10 = 4
x = -2
Thus, x = -2 and y = 5
Now, using the Cross-multiplication method,
8x +5y – 9 = 0
3x + 2y – 4 = 0
$\frac{x}{-20+18} = \frac{y}{-27+32} = \frac{1}{16-15}$

$\frac{-x}{2} = \frac{y}{5} =\frac{1}{1}$
∴ x = -2 and y =5

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method.

(i) A part of monthly hostel charges is fixed, and the remaining depends on the number of days one has taken food in a mess. When student A takes food for 20 days, she has to pay Rs.1,000 as hostel charges, whereas student B, who takes food for 26 days, pays Rs.1,180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator, and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks on a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and its breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solutions –

(i) Let x be the fixed charge of the food and y be the charge for food per day.
According to the question,
x + 20y = 1000 ……………….. (i)
x + 26y = 1180 ……………….. (ii)
Subtracting (i) from (ii), we get
(x + 26y) – (x + 20y) = 1180 – 1000
6y = 180
y = Rs.30
Using this value in equation (ii), we get
x = 1180 – 26 × 30
x = 400
Therefore, the fixed charge is Rs.400, and the charge per day is Rs.30.

(ii) Let the fraction be x/y.
So, as per the question given,
$\frac{(x-1)}{y}$ = $\frac{1}{3}$
3x – y = 3 ………………… (i)

$\frac{x}{(y+8)}$ = $\frac{1}{4}$
4x – y = 8 ……………….. (ii)
Subtracting equation (i) from (ii), we get
(4x – y) – (3x – y) = 8 – 3
x = 5 ………………. (iii)
Using this value in equation (ii), we get
(4 × 5) – y = 8
y = 12
Therefore, the fraction is 5/12.

(iii) Let the number of right answers be x and the number of wrong answers be y.
According to the given question,
3x − y = 40 …….. (i)
4x − 2y = 50
2x − y = 25 ……. (ii)
Subtracting equation (ii) from equation (i), we get
(3x − y) – (2x − y) = 40 – 25
x = 15 ….…. (iii)
Putting this in equation (ii), we obtain
30 – y = 25
y = 5
Therefore, number of right answers = 15 and number of wrong answers = 5
Hence, the total number of questions = 20

(iv) Let x km/h be the speed of the car from point A and y km/h be the speed of the car from point B.
If the car travels in the same direction,
5x – 5y = 100
x – y = 20 ……………… (i)
If the car travels in the opposite direction,
x + y = 100 ……………… (ii)
Solving equations (i) and (ii), we get
(x – y) + (x + y) = 20 + 100
2x = 120
x = 60 …………………… (iii)
Using this in equation (i), we get
60 – y = 20
y = 40
Therefore, the speed of the car from point A = 60 km/h
Speed of car from point B = 40 km/h

(v) Let, The length of the rectangle = x unit
And the breadth of the rectangle = y unit
Now, as per the question given,
(x – 5) (y + 3) = xy – 9
xy + 3x – 5y – 15 = xy – 9
3x – 5y – 6 = 0 ……………… (i)
(x + 3) (y + 2) = xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y – 61 = 0 …………….. (ii)
Using cross multiplication method, we get
$\frac{x}{305+18}= \frac{y}{-12+183}=\frac{1}{9+10}$

$\frac{x}{323} = \frac{y}{171} = \frac{1}{19}$

$\frac{x}{323} =\frac{1}{19}$ and $\frac{y}{171} =\frac{1}{19}$
Therefore, x = 17 and y = 9
Hence, the length of the rectangle = 17 units
And the breadth of the rectangle = 9 units

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NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5