Answer:
1) Exactly one solution
2) Exactly one solution
3) No solution
4) Infinite many solution
5) No solution
Step-by-step explanation:
As given
1)
y = 3x + 1 ......(1)
y = -3x + 7 ......(2)
Put the value of y in equation (1) , we get
-3x + 7 = 3x + 1
⇒7 - 1 = 3x + 3x
⇒6 = 6x
⇒x = 1
Now as
y = -3x + 7
⇒y = -3(1) + 7 = -3 + 7 = 4
∴ we get
x = 1, y = 4
Unique Solution ( Exactly one solution )
2)
y = 3x+ 1 ......(3)
y = x+1 .......(4)
Put the value of y in equation (3), we get
x + 1 = 3x + 1
⇒1 - 1 = 3x - x
⇒0 = 2x
⇒x = 0
Now, as
y = 3x + 1
⇒y = 3(0) + 1 = 1
∴ we get
x = 0, y = 1
Unique solution ( Exactly one solution )
3)
y = 3x + 1 ........(5)
y = 3x + 7 ........(6)
Put the value of y in equation (5), we get
3x + 7 = 3x + 1
⇒7 - 1 = 3x - 3x
⇒6 = 0
No Solution
4)
x+y=10 ........(7)
2x+2y=20 .........(8)
Divide equation (8) by 2, we get
x + y = 10 .....(9)
⇒y = 10 - x
Put the value of y in equation (7), we get
x + 10 - x = 10
⇒10 = 10
Infinite many solutions
5)
x + y = 10 .......(10)
x + y = 12 .......(11)
⇒y = 12 - x
Put the value of y in equation (10) , we get
x + 12 - x = 10
⇒12 = 10
No solution
