# Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5

### Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

### (i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

**Solution:**

Here,

a

_{1}= 1, b_{1}= -3, c_{1}= -3a

_{2}= 3, b_{2}= -9, c_{2}= -2So,

As,

Hence, the given pairs of equations have

no solution.

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

**Solution:**

Rearranging equations, we get

2x + y -5 = 0

3x + 2y -8 = 0

Here,

a

_{1}= 2, b_{1}= 1, c_{1}= -5a

_{2}= 3, b_{2}= 2, c_{2}= -8So,

As,

Hence, the given pairs of equations have

unique solution.For cross multiplication,

= y = 1

Hence,

= 1

x = 2and,

y = 1Hence, the

required solution is x = 2 and y = 1.

### (iii) 3x – 5y = 20, 6x – 10y = 40

**Solution:**

Rearranging equations, we get

3x – 5y – 20 = 0

6x – 10y – 40 = 0

Here,

a

_{1}= 3, b1 = -5, c1 = -20a

_{2}= 6, b_{2}= -10, c_{2}= -40So,

As,

Hence, the given pairs of equations have

infinitely many solutions.

### (iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

**Solution:**

Here,

a

_{1}= 1, b1 = -3, c1 = -7a

_{2}= 3, b_{2}= -3, c_{2}= -15So,

As,

Hence, the given pairs of equations have

unique solution.For cross multiplication,

Hence,

x =

x = 4and, y =

y = -1Hence, the

required solution is x = 4 and y = -1.

### Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

### 2x + 3y = 7

### (a – b) x + (a + b) y = 3a + b – 2

**Solution:**

Here,

a

_{1}= 2, b1 = 3, c1 = -7a

_{2}= a-b, b_{2}= a+b, c_{2}= -(3a+b-2)For having infinite number of solutions, it has to satisfy below conditions:

Now, on comparing

2(a+b) = 3(a-b)

2a+2b = 3a – 3b

a – 5b = 0

…………………(1)And, now on comparing

3(3a+b-2) = 7(a+b)

9a+3b-6 = 7a+7b

2a-4b-6=0

Reducing form, we get

a-2b-3=0

…………………(2)Now, new values for

a

_{1}= 1, b_{1}= -5, c_{1}= 0a

_{2}= 1, b_{2}= -2, c_{2}= -3

Solving Eq(1) and Eq(2), by cross multiplication,Hence,

a =

a = 5and, b =

b = 1Hence, For values

a = 5 and b = 1pair of linear equations have an infinite number of solutions

### (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:**

Here,

a

_{1}= 3, b_{1}= 1, c_{1}= -1a

_{2}= (2k-1), b_{2}= k-1, c_{2}= -(2k+1)For having no solution, it has to satisfy below conditions:

Now, on comparing

3(k-1) = 2k-1

3k-3 = 2k-1

k = 2And, now on comparing

2k+1 ≠ k-1

k ≠ -2Hence, for

k = 2 and k ≠ -2the pair of linear equations have no solution.

### Question 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 = 0 …………………(1)

3x + 2y – 4 = 0 …………………(2)

Substitution MethodFrom Eq(2), we get

x = 4-2y/3

Now, substituting it in Eq(1), we get

8(4-2y/3) + 5y – 9 = 0

32-16y/3 + 5y – 9 = 0

32 – 16y + 15y – 27 = 0

y = 5Now, substituting y = 5 in Eq(2), we get

3x + 2(5) – 4 = 0

3x = -6

x = -2

Cross Multiplication MethodHere,

a

_{1}= 8, b_{1}= 5, c_{1}= -9a

_{2}= 3, b_{2}= 2, c_{2}= -4So,

As,

Hence, the given pairs of equations have unique solution.

For cross multiplication,

= 1

Hence,

= 1

x = -2and, = 1

y = 5Hence,

the required solution is x = -2 and y = 5.

### Question 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 the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

**Solution:**

Let’s take,

Fixed charge = x

Charge of food per day = y

According to the given question,

x + 20y = 1000

……………….. (1)x + 26y = 1180

………………..(2)Subtracting Eq(1) from Eq(2) we get

6y = 180

y = 30Now, substituting y = 30 in Eq(2), we get

x + 20(30) = 1000

x = 1000 – 600

x= 400.Hence,

fixed charges is ₹ 400 and charge per day is ₹ 30.

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

**Solution:**

Let the fraction be .

So, as per the question given,

3x – y = 3

…………………(1)4x –y =8

………………..(2)Subtracting Eq(1) from Eq(2) , we get

x = 5Now, substituting x = 5 in Eq(2), we get

4(5)– y = 8

y = 12Hence, the fraction is .

### (iii) Yash scored 40 marks in 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?

**Solution:**

Let’s take

Number of right answers = x

Number of wrong answers = y

According to the given question;

3x−y=40

……….……..(1)4x−2y=50

2x−y=25

……………….(2)Subtracting Eq(2) from Eq(1), we get

x = 15Now, substituting x = 15 in Eq(2), we get

2(15) – y = 25

y = 30-25

y = 5Hence, number of right answers = 15 and number of wrong answers = 5

Hence,

total number of questions = 20

### (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?

**Solution:**

Let’s take

Speed of car from point A = x km/he

Speed of car from point B = y km/h

When car travels in the same direction,

5x – 5y = 100

x – y = 20

………………(1)When car travels in the opposite direction,

x + y = 100

………………..(2)Subtracting Eq(1) from Eq(2), we get

2y = 80

y = 40Now, substituting y = 40 in Eq(1), we get

x – 40 = 20

x = 60Hence, the speed of car

from point A = 60 km/hSpeed of car

from point B = 40 km/h.

### (v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and 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.

**Solution:**

Let’s take

Length of rectangle = x unit

Breadth of the rectangle = y unit

Area of rectangle will be = xy sq. units

According to the given conditions,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0

……………………(1)(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0

……………………..(2)Using cross multiplication method, we get,

Hence,

x = 17and,

y = 9Hence, the required solution is

x = 17 and y = 9.

Length of rectangle = 17 units

Breadth of the rectangle = 9 units