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.
Question 1.
Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 β¦(i)
2x-3y=12 β¦(ii)
Solution:
5x + 3y = 9 β¦(i)
2x-3y=12 β¦(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 β 2y = 7
iii. 2x β 3y = 9; 2x + y = 13
iv. 5m β 3n = 19; m β 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 β 57y = 172; 57x β 49y = 252
Solution:
i. 3a + 5b = 26 β¦(i)
a + 5b = 22 β¦(ii)
Subtracting equation (ii) from (i), we get

Substituting a = 2 in equation (ii), we get
2 + 5b = 22
β΄ 5b = 22 β 2
β΄ 5b = 20
β΄ b = \(\frac { 20 }{ 5 } \) =4
β΄ (a, b) = (2, 4) is the solution of the given simultaneous equations.
ii. x + 7y = 10
β΄ x = 10 β 7y β¦(i)
3x β 2y = 7 β¦1(ii)
Substituting x = 10 β ly in equation (ii), we get
3 (10 β 7y) β 2y = 7
β΄ 30 β 21y β 2y = 7
β΄ -23y = 7 β 30
β΄ -23y = -23
β΄ y = \(\frac { -23 }{ -23 } \)
Substituting y = 1 in equation (i), we get
x = 10 β 7 (1)
= 10 β 7 = 3
β΄ (x, y) = (3, 1) is the solution of the given simultaneous equations.
iii. 2x β 3y = 9 β¦(i)
2x + y = 13 β¦(ii)
Subtracting equation (ii) from (i), we get

β΄ (x, y) = (6, 1) is the solution of the given simultaneous equations.
iv. 5m β 3n = 19 β¦(i)
m β 6n = -7
β΄ m = 6n β 7 β¦(ii)
Substituting m = 6n β 7 in equation (i), we get
5(6n β 7) β 3n = 19
β΄ 30n β 35 β 3n = 19
β΄ 27n = 19 + 35
β΄ 27n = 54
β΄ n = \(\frac { 54 }{ 27 } \) = 2
Substituting n = 2 in equation (ii), we get
m = 6(2) β 7
= 12 β 7 = 5
β΄ (m, n) = (5, 2) is the solution of the given simultaneous equations.
v. 5x + 2y = -3 β¦(i)
x + 5y = 4
β΄ x = 4 β 5y β¦(ii)
Substituting x = 4 β 5y in equation (i), we get
5(4 β 5y) + 2y = -3
β΄ 20 β 25y + 2y = -3
β΄ -23y = -3 β 20
β΄ -23y = -23
β΄ y = \(\frac { -23 }{ -23 } \) = 1
Substituting y = 1 in equation (ii), we get
x = 4 β 5(1)
= 4 β 5 = -1
β΄ (x, y) = (-1, 1) is the solution of the given simultaneous equations.

Substituting y = 3 in equation (i), we get
x = 10 β 3(3)
= 10 β 9 = 1
β΄ (x, y) = (1, 3) is the solution of the given simultaneous equations.
vii. 99x + 101 y = 499 β¦(i)
101 x + 99y = 501 β¦(ii)
Adding equations (i) and (ii), we get

Substituting x = 3 in equation (iii), we get
3 + y = 5
β΄ y = 5 β 3 = 2
β΄ (x, y) = (3, 2) is the solution of the given simultaneous equations.
viii. 49x β 57y = 172 β¦(i)
57x β 49y = 252 β¦(ii)
Adding equations (i) and (ii), we get

Substituting x = 7 in equation (iv), we get
7 + y = 10
β΄ y = 10 β 7 = 3
β΄ (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 β y = 18 (Textbook pg. no. 3)
Solution:
3x + 2y = 29 β¦(i)
and 5x- y = 18 β¦(ii)
Letβs solve the equations by eliminating βyβ.
Fill suitably the boxes below.
Multiplying equation (ii) by 2, we get