Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths 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

(ii) 2x + y = 5

3x + 2y = 8

(iii) 3x – 5y = 20

6x – 10y = 40

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

Solution:

(i) The given pair of linear equations is

x – 3y – 3 = 0

and 3x – 9y – 2 = 0

Here a_{1} = 1, b_{1} = – 3, c_{1} = – 3

a_{2} = 3, b_{2} = – 9, c_{2} = – 2

Hence the given pair of linear equations has no solution.

(ii) The given pair of linear equations is

2x + y = 5

and 3x + 2y = 8

or 2x + y – 5 = 0

and 3x + 2y – 8 = 0

Here a_{1} = 2, b_{1} = 1, c_{1}= – 5

a_{2} = 3, b_{2} = 2, c_{2} = – 8

∴ The given pair of equations has a unique solution.

Then solving by cross-multiplication method,

Hence, x = 2 and y = 1

(iii) The given pair of linear equations is

3x – 5y – 20

and 6x – 10y = 40

or 3x – 5y – 20 = 0

and 6x – 10y – 40 = 0

Here a_{1} = 3, b_{1} = -5, c_{1} = -20

a_{2} = 6, b_{2} = -10, c_{2} = -40

Hence the given pair of equations has infinitely many solutions.

(iv) The given pair of linear equations is

x – 3y – 7 = 0

and 3x – 3y – 15 = 0

Here a_{1} = 1, b_{1} = -3, c_{1} = – 7

a_{2} = 3, b_{2} = – 3, c_{2} = – 15

∴ The given pair of equations has a unique solution.

Now solving by cross-multiplication method,

Hence, x = 2 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

(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) The given pair of linear equations is

2x + 3y = 7

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

or 2x + 3y – 7 = 0

and (a – b) x + (a + b) y – (3a + b – 2) = 0

Here a_{1} = 2, b_{1} = 3, c_{1} = – 7

a_{2} = a – b, b_{2} = a + b, c_{2} = -(3a + b – 2)

∴ The given pair of equations has infinitely many solutions

or 6a + 2b – 4 = 7a – 7b

or – a + 9b – 4= 0

or a = 9b – 4 ….(1)

From II and III,

\(\frac{3}{a+b}\) = \(\frac{7}{3 a+b-2}\)

or 9a + 3b – 6= 7a + 7b

or 2a – 4b – 6 = 0

or a – 2b – 3 = 0 ….(2)

Substituting the value of a from equation (1) in (2) :

9b – 4 – 2b – 3 = 0

or 7b – 1 = 9

or 7b = 1

or b = 1

Substituting this value of b in (1)

a – 9 × 1 – 4 = 9 – 4

⇒ a = 5

Hence the required solution of the pair of linear equations is a = 5 and b = 1.

(ii) The given pair of linear equations is 3x + y = 1

and (2k – 1) x + (k – 1) y = 2k+ 1

or 3x + y – 1 = 0

and (2k – 1) x + (k – 1) y – (2k + 1) = 0

Here a_{1} = 3, b_{1} = 1, c_{1} = – 1

a_{2} = (2k – 1), b_{2} = k – 1, c_{2} = – (2k + 1)

∴ The given pair of equations has no solution

⇒ 3k – 2k = – 1 + 3

⇒ k = 2

For k = 2 it is clear that

\(\frac{1}{k-1}\) ≠ \(\frac{1}{2 k+1}\)

Hence for k = 2 the given pair of linear equations has 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:

The given pair of linear equations is

8x + 5y = 9 ……..(1)

3x + 2y = 4 ……(2)

By Substitution Method—

From (2), 2y= 4 – 3x

⇒ y = \(\frac{4-3 x}{2}\) …(3)

Substituting this value of y in equation (1),

8x + 5\(\left[\frac{4-3 x}{2}\right]\) = 9

or \(\frac{16 x+20-15 x}{2}\) = 9

or x + 20 = 18

or x = 18 – 20 = – 2

Substituting this value of x in equation (3),

y = \(\frac{4-3(-2)}{2}\) = \(\frac{4+6}{2}\)

= \(\frac{10}{2}\) = 5

Hence, x = – 2 and y = 5

By Cross-multiplication Method—

The pair of linear equations is

8x + 5y – 9 = 0

and 3x + 2y – 4 = 0

Here a_{1} = 8, b_{1} = 5, c_{1} = – 9

a_{2} = 3, b_{2} = 2, c_{2} = – 4

Hence the solution of the given pair of linear equations 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 Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

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

(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?

(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 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:

(i) Let the fixed charges of the hostel = Rs. x

and the cost of food per day = Rs. y

According to the first condition,

x + 20y = 1000 …. (1)

According to the second condition,

x + 26y = 1180 ….(2)

Hence the fixed charges of the hostel and the cost of food per day are Rs. 400 and Rs. 30 respectively.

(ii) Let the numerator of the fraction = x

and the denominator the fraction = y

∴ Required fraction = y

According to the first condition,

\(\frac{x-1}{y}\) = \(\frac{1}{3}\)

⇒ 3x – 3 = y

⇒ 3x – y – 3 = 0 …..(1)

According to the second condition,

\(\frac{x}{y+8}\) = \(\frac{1}{4}\)

or 4x – y + 8

or 4x – y – 8 = 0 …. (2)

Hence, the required fraction is \(\frac{5}{12}\).

(iii) Let the number of questions answered rightly by Yash = A

and the number of questions answered wrong by Yash = y

According to the first condition,

3x – y = 40

⇒ 3x – y – 40 = 0 ……(1)

According to the second condition,

4x – 2y = 50

⇒ 4x – 2y – 50 = 0 …….(2)

Then, solving by cross-multiplication method,

or y = 4

∴ Number of right questions = 15

and Number of wrong questions = 5

Hence total number of questions

= [Number of right questions] + [Number of wrong questions]

= 15 + 5 = 20

(iv) Let the speed of the car starting from place A = x km/hour

and the speed of the car starting from place B = y km/hour

Distance between A and B = 100 km.

In case of 5 hours—

Distance covered by car A = 5x km.

[∵ Distance = Speed × Time]

Distance covered by car B = 5y km.

According to the first condition,

5x – 5y = 100

or x – y = 20

(∵ Cars travel in the same direction)

or x – y – 20 = 0 …. (1)

In case of 1 hour—

Distance covered by car A = A km.

[∵ Distance = Speed × Time]

Distance covered by car B = y km.

According to the second condition,

x + y = 100

or x + y – 100 = 0 …(2)

Hence the speeds of the cars starting from palces A and B are respectively 60 km/hr and 40 km/hr.

(v) Let the length of the rectangle = x units

and the breadth of the rectangle = y units

∴ Area of the rectangle = xy sq. units

According to the first condition,

(x – 5) (y + 3) = xy – 9

or xy + 3x – 5y – 15 = xy – 9

or 3x – 5y – 6 = 0 …. (1)

According to the second condition,

(x + 3) (y + 2) = xy + 67

xy + 2x + 3y + 6 = xy + 67

or 2x + 3y – 61 = 0 …(2)

Hence the length and the breadth of the rectangle are 17 units and 9 units respectively.

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