(i) 2x+y = 7
Solution:
To find the four solutions of 2x+y =7 we substitute different values for x and y
Let x = 0
Then,
2x+y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Then,
2x+y = 7
(2×1)+y = 7
2+y = 7
y = 7-2
y = 5
(1,5)
Let y = 1
Then,
2x+y = 7
(2x)+1 = 7
2x = 7-1
2x = 6
x = 6/2
x = 3
(3,1)
Let x = 2
Then,
2x+y = 7
(2×2)+y = 7
4+y = 7
y =7-4
y = 3
(2,3)
The solutions are (0, 7), (1,5), (3,1), (2,3)
(ii) πx+y = 9
Solution:
To find the four solutions of πx+y = 9 we substitute different values for x and y
Let x = 0
Then,
Ï€x+y = 9
(Ï€×0)+y = 9
y = 9
(0,9)
Let x = 1
Then,
Ï€x +y = 9
(Ï€×1)+y = 9
Ï€+y = 9
y = 9-Ï€
(1, 9-Ï€)
Let y = 0
Then,
Ï€x+y = 9
Ï€x+0 = 9
Ï€x = 9
x = 9/Ï€
(9/Ï€,0)
Let x = -1
Then,
Ï€x + y = 9
(Ï€×-1) + y = 9
-Ï€+y = 9
y = 9+Ï€
(-1,9+Ï€)
The solutions are (0,9), (1,9-Ï€), (9/Ï€,0), (-1,9+Ï€)
(iii) x = 4y
Solution:
To find the four solutions of x = 4y we substitute different values for x and y
Let x = 0
Then,
x = 4y
0 = 4y
4y= 0
y = 0/4
y = 0
(0,0)
Let x = 1
Then,
x = 4y
1 = 4y
4y = 1
y = 1/4
(1,1/4)
Let y = 4
Then,
x = 4y
x= 4×4
x = 16
(16,4)
Let y = 1
Then,
x = 4y
x = 4×1
x = 4
(4,1)
The solutions are (0,0), (1,1/4), (16,4), (4,1)
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