## Tamilnadu State Board Class 10 Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.

Solve the following system of linear equations in three variables

(i) x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16

(ii) \(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) + 4 = 0; \(\frac { 1 }{ y } \) – \(\frac { 1 }{ z } \) + 1 = 0; \(\frac { 2 }{ z } \) + \(\frac { 3 }{ x } \) = 14

(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)

Solutions:

(i) x + y + z = 5 …(1)

2x – y + z = 9 …(2)

x – 2y + 3z = 16 …(3)

Substitute z = 4 in (4)

3x + 2(4) = 14

3x + 8 = 14

3x = 6

x = 2

Substitute x = 2, z = 4 in (1)

2 + y + 4 = 5 ⇒ y = -1

x = 2, ,y = -1, z = 4

a – 2b + 4 = 0 ⇒ a – 2b = -4 …(1)

b – c + 1 = 0 ⇒ b – c = -1 …(2)

2c + 3a = 14 ⇒ 2c + 3a = 14 …(3)

(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)

x = \(\frac { 3y }{ 2 } \) – 10 …..(1)

2z + 5 = 110 – (y + z)

2z = 105 – y – z

y = 105 – 3z …..(2)

Substitute (2) in (1), x = \(\frac { 315 }{ 2 } \) – \(\frac { 9z }{ 2 } \) – 10

= 2z + 5 – 20

∴ 315 – 9z – 20 = 4z – 30

13 z = 315 – 20 + 30

= 325

z = \(\frac { 325 }{ 13 } \) = 25

x + 20 = 2z + 5

x + 20 = 50 + 5

x = 35

Substitute z = 25 in (2)

y = 105 – 3z = 105 – 75 = 30

y = 30

x = 35, y = 30, z = 25

The system has unique solutions.

Question 2.

Discuss the nature of solutions of the following system of equations

(i) x + 2y – z = 6 ; -3x – 2y + 5z = -12 ; x – 2z = 3

(ii) 2y + z = 3 (-x + 1); -x + 3y -z = -4 3x + 2y + z = – \(\frac { 1 }{ 2 } \)

(iii) \(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \); x + y + z = 27

Solution:

(i) x + 2y – z = 6 …(1)

-3x – 2y + 5z = -12 …(2)

x – 2z = 3 …(3)

We see that the system has an infinite number of solutions.

(ii) 2y + z = 3(-x + 1);

-x + 3y – z = -4;

3x + 2y + z = –\(\frac { 1 }{ 2 } \)

2y + z + 3x = 3 ⇒ 3x + 2y + z = 3 ….(1)

-x + 3y – z = -4 ….(2)

3x + 2y + z = –\(\frac { 1 }{ 2 } \) ….(3)

This is contradiction. This means the system is inconsistent and has no solutions.

∴ The system has unique solutions.

Question 3.

Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now ?

Solution:

Let Vani’s age be x

Let Vani’s father’s age be y

Let Vani’s grand father’s age be z.

Sub, z = 84 in (3), we get

4x – 84 = 12

4x = 96

x = 24

Sub, x = 24, z = 84 in (1) we get

24 + y + 84 = 159

y = 159 – 108

= 51

∴ Vani’s age = 24 years

Her father’s age =51 years

Her grand father’s age = 84 years.

Question 4.

The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?

Solution:

Let the number be 100x + 10y + z.

Reversed number be 100z + 10y + x.

x + y + z = 11 …(1)

100z + 10y + x = 5(100x + 10y + z) + 46

100z + 10y + x = 500x + 50y + 5z + 46

499x + 40y – 95z -46 …(2)

x + 2y = z

x + 2y – z = 0 ….(3)

Question 5.

There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.

Solution:

Let x, y and z be number of currency pieces of 5,10,20 rupees

x + y + z = 12 …(1)

5x + 10y + 20z = 105 …(2)

10x + 5y + 20z = 125 …(3)

Sub, z = 2 in (5), we get

15y + 20 × 2 = 85

15y = 45

y = 3

Sub; y = 3, z = 2 in (1)

x + y + z = 12

x = 7

∴ The solutions are

the number of ₹ 5 are 7

the number of ₹ 10 are 3

the number of ₹ 20 are 2