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
So, x =^1221
12 21
bc b c
ab a b, provided a 1 b 2 – a 2 b 1 ✁ 0 (5)Step 3 : Substituting this value of x in (1) or (2), we get
y =^1221
12 21ca c a
ab a b(6)
Now, two cases arise :
Case 1 : a 1 b 2 – a 2 b 1 ✁ 0. In this case^11
22
ab
ab✂. Then the pair of linear equations hasa unique solution.
Case 2 : a 1 b 2 – a 2 b 1 = 0. If we write^11
22
abk
ab✄ ✄ , then a 1 = k^ a 2 , b 1 = k^ b 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 ifc 1 = k c 2 , i.e.,^1
2
c k.
c✄
If c 1 = k c 2 , any solution of Equation (2) will satisfy the Equation (1), and viceversa. So, if^111
222
abck
abc☎ ☎ ☎ , then there are infinitely many solutions to the pair oflinear 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:
(i) When^11
22ab
ab✂ , we get a unique solution.(ii)When^111
222abc
abc☎ ☎ , there are infinitely many solutions.(iii) When^111
222abc
abc✄ ✂ , there is no solution.