Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

## Practice Set 1.1 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.

Complete the following activity to solve the simultaneous equations.

5x + 3y = 9 …(i)

2x-3y=12 …(ii)

Solution:

5x + 3y = 9 …(i)

2x-3y=12 …(ii)

Add equations (i) and (ii).

Question 2.

Solve the following simultaneous equations.

i. 3a + 5b = 26; a + 5b = 22

ii. x + 7y = 10; 3x – 2y = 7

iii. 2x – 3y = 9; 2x + y = 13

iv. 5m – 3n = 19; m – 6n = -7

v. 5x + 2y = -3;x + 5y = 4

vi. \(\frac { 1 }{ 3 } \) x+ y = \(\frac { 10 }{ 3 } \) ; 2x + \(\frac { 1 }{ 4 } \) y = \(\frac { 11 }{ 4 } \)

vii. 99x + 101y = 499 ; 101x + 99y = 501

viii. 49x – 57y = 172; 57x – 49y = 252

Solution:

i. 3a + 5b = 26 …(i)

a + 5b = 22 …(ii)

Subtracting equation (ii) from (i), we get

Substituting a = 2 in equation (ii), we get

2 + 5b = 22

∴ 5b = 22 – 2

∴ 5b = 20

∴ b = \(\frac { 20 }{ 5 } \) =4

∴ (a, b) = (2, 4) is the solution of the given simultaneous equations.

ii. x + 7y = 10

∴ x = 10 – 7y …(i)

3x – 2y = 7 …1(ii)

Substituting x = 10 – ly in equation (ii), we get

3 (10 – 7y) – 2y = 7

∴ 30 – 21y – 2y = 7

∴ -23y = 7 – 30

∴ -23y = -23

∴ y = \(\frac { -23 }{ -23 } \)

Substituting y = 1 in equation (i), we get

x = 10 – 7 (1)

= 10 – 7 = 3

∴ (x, y) = (3, 1) is the solution of the given simultaneous equations.

iii. 2x – 3y = 9 …(i)

2x + y = 13 …(ii)

Subtracting equation (ii) from (i), we get

∴ (x, y) = (6, 1) is the solution of the given simultaneous equations.

iv. 5m – 3n = 19 …(i)

m – 6n = -7

∴ m = 6n – 7 …(ii)

Substituting m = 6n – 7 in equation (i), we get

5(6n – 7) – 3n = 19

∴ 30n – 35 – 3n = 19

∴ 27n = 19 + 35

∴ 27n = 54

∴ n = \(\frac { 54 }{ 27 } \) = 2

Substituting n = 2 in equation (ii), we get

m = 6(2) – 7

= 12 – 7 = 5

∴ (m, n) = (5, 2) is the solution of the given simultaneous equations.

v. 5x + 2y = -3 …(i)

x + 5y = 4

∴ x = 4 – 5y …(ii)

Substituting x = 4 – 5y in equation (i), we get

5(4 – 5y) + 2y = -3

∴ 20 – 25y + 2y = -3

∴ -23y = -3 – 20

∴ -23y = -23

∴ y = \(\frac { -23 }{ -23 } \) = 1

Substituting y = 1 in equation (ii), we get

x = 4 – 5(1)

= 4 – 5 = -1

∴ (x, y) = (-1, 1) is the solution of the given simultaneous equations.

Substituting y = 3 in equation (i), we get

x = 10 – 3(3)

= 10 – 9 = 1

∴ (x, y) = (1, 3) is the solution of the given simultaneous equations.

vii. 99x + 101 y = 499 …(i)

101 x + 99y = 501 …(ii)

Adding equations (i) and (ii), we get

Substituting x = 3 in equation (iii), we get

3 + y = 5

∴ y = 5 – 3 = 2

∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

viii. 49x – 57y = 172 …(i)

57x – 49y = 252 …(ii)

Adding equations (i) and (ii), we get

Substituting x = 7 in equation (iv), we get

7 + y = 10

∴ y = 10 – 7 = 3

∴ (x, y) = (7, 3) is the solution of the given simultaneous equations.

Complete the following table. (Textbook pg. no. 1)

Question 1.

Solve: 3x+ 2y = 29; 5x – y = 18 (Textbook pg. no. 3)

Solution:

3x + 2y = 29 …(i)

and 5x- y = 18 …(ii)

Let’s solve the equations by eliminating ‘y’.

Fill suitably the boxes below.

Multiplying equation (ii) by 2, we get