10th Standard Maths 1 Practice Set 1.1 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board
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.
Class 10 Maths Part 1 Practice Set 1.1 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board
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
 
Maharashtra State Board Class 10 Maths Solutions Part 1
- Linear Equations in Two Variables Practice Set 1.1 Class 10 Maths Solutions
 - Linear Equations in Two Variables Practice Set 1.2 Class 10 Maths Solutions
 - Linear Equations in Two Variables Practice Set 1.3 Class 10 Maths Solutions
 - Linear Equations in Two Variables Practice Set 1.4 Class 10 Maths Solutions
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