Class 10 Maths Part 1 Practice Set 1.1 Solutions – Chapter 1 Linear Equations in Two Variables | Maharashtra Board

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Class 10 Maths Part 1 Practice Set 1.1 Solutions – Chapter 1 Linear Equations in Two Variables | Maharashtra Board

Class 10 Maths Part 1 Practice Set 1.1 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board Practice Set 1.1 Que

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Class 10 Maths Part 1 Practice Set 1.1 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Practice Set 1.1

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).

Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 1

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. 1/3 x+ y = 10/3 ; 2x + 1/4 y = 11/4
vii. 99x + 101y = 499 ; 101x + 99y = 501

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

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

Solution:
i. 3a + 5b = 26 …(i)
a + 5b = 22 …(ii)
Subtracting equation (ii) from (i), we get

Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 2

Substituting a = 2 in equation (ii), we get
2 + 5b = 22
∴ 5b = 22 – 2
∴ 5b = 20
∴ b = 20/5 =4
∴ (a, b) = (2, 4) is the solution of the given simultaneous equations.

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ii. x + 7y = 10; 3x – 2y = 7

Solution:
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 = 2323
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; 2x + y = 13
Solution:
2x – 3y = 9 …(i)
2x + y = 13 …(ii)
Subtracting equation (ii) from (i), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 3
∴ (x, y) = (6, 1) is the solution of the given simultaneous equations.
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iv. 5m – 3n = 19; m – 6n = -7
Solution:
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 = 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.
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vi. 13 x+ y = 103 ; 2x + 14 y = 114
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 4
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vii. 99x + 101y = 499 ; 101x + 99y = 501
Solution:
99x + 101 y = 499 …(i)
101 x + 99y = 501 …(ii)
Adding equations (i) and (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 5
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.
—————————————————————
vii. 99x + 101y = 499 ; 101x + 99y = 501
Solution:
49x – 57y = 172 …(i)
57x – 49y = 252 …(ii)
Adding equations (i) and (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 6
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.

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