Maths

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 59
i.e., (b 2
a 1
– b 1
a 2
) x = b 1
c 2
– b 2
c 1
bc
So, x = 1 2 b 2 c
1
ab a b , provided a b – a b 0 (5)
1 2 2 1 ✁
1 2 2 1
Step 3 : Substituting this value of x in (1) or (2), we get
ca
y = 1 2 ca
2 1
ab ab
1 2 2 1
Now, two cases arise :
a1 b1
Case 1 ✁ : a 1
b 2
– ✂
a 2
b 1
0. In this case . Then the pair of linear equations has
a2 b2
a unique solution.
a1 b1
Case 2 : a 1
b 2
– a 2
b 1
= 0. If we write k
a b
, then a = k a , b = k b .
1 2 1 2 ✄ ✄
2 2
Substituting the values of a 1
and b 1
in the Equation (1), we get
k (a 2
x + b 2
y) + c 1
= 0. (7)
It can be observed that the Equations (7) and (2) can both be satisfied only if
c 1
= k c 2
, i.e.,
c
1
c ✄
2
k.
If c 1
= k c 2
, any solution of Equation (2) will satisfy the Equation (1), and vice
a1 b1 c1
versa. So, if k
a b c
, then there are infinitely many solutions to the pair of
☎ ☎ ☎
2 2 2
linear equations given by (1) and (2).
If c 1
✁ k c 2
, then any solution of Equation (1) will not satisfy Equation (2) and vice
versa. Therefore the pair has no solution.
We can summarise the discussion above for the pair of linear equations given by
(1) and (2) as follows:
a1 b1
✂
(i) When , we get a unique solution.
a b
(ii) When
2 2
a1 b1 c1
☎ ☎ , there are infinitely many solutions.
a b c
2 2 2
a1 b1 c1
✄ (iii) ✂ When , there is no solution.
a b c
2 2 2
(6)