1. Linear Equations in Two Variables
Practice Set 1.2
1.Linear Equations in Two Variables
1. Complete the following table to draw graph of the equations
(I) x + y = 3
(II) x – y = 4
Solution:
(x+y=3)
(x, y) = (3, 0) and x = 3 ……..(Given)
∴y = 0
Substituting
y = 5
x + 5 = 3
∴x = – 2
(x, y) = (0, 3) and y = 3 (Given)
∴x = 0
(x-y=4)
Substituting y = 0
x – 0 = 4
∴x = 4
Substituting x = -1
– 1 – y = 4
∴y = – 5
2. Solve the following simultaneous equations graphically.
(1) x + y = 6 ; x – y = 4
Solution:
x+y=6
x-y=4
(2) x + y = 5 ; x – y = 3
Solution:
x+y=5 ∴∴y=5-x
x-y=3∴ y=x-3x-3
The coordinates of the point of intersection are (4, 1).
Ans. The solution of the given simultaneous equations is x = 4 , y = 1 .
(3) x + y = 0 ; 2x – y = 9
Solution:
x + y = 0
∴y = – x
2x – y = 9
∴y = 2x – 9
The coordinates of the point of intersection are (3, -3)
Ans. The solution of the given simultaneous equations is x = 3 , y = -3 .
(4) 3x – y = 2 ; 2x – y = 3
Solution:
3x – y = 2
∴3x – 2 = y
∴y = 3x – 2
2x – y = 3
∴y = 2x – 3
The coordinates of the point of intersection
are (-1,-5).
Ans. The solution of the given simultaneous equations is x = – 1, y = – 5 .
(5) 3x – 4y = -7 ; 5x – 2y = 0
Solution:
3x – 4y = – 7
∴3x + 7 = 4y
∴
5x – 2y = 0
∴5x = 2y
∴
The coordinates of the point of intersection
are (1, 2.5).
Ans. The solution of the given simultaneous equations is x = 1, y = 2.5 .
(6) 2x – 3y = 4 ; 3y – x = 4
Solution:
2x – 3y = 4
∴2x – 4 = 3y
3y-x-4
∴3y = x + 4
The coordinates of the point of intersection are (8, 4).
Ans. The solution of the given simultaneous equations is x = 8 , y = 4
1.Linear Equations in Two Variables
