58 MATHEMATICS
Therefore, x =
– (4)(–35) (3)( 28)✁
(5)(4) (3)(2)
✂ ✂
✂
i.e., x =
(3) (– 28) (4) ( 35)
(5)(4) (2)(3)
✄ ✄
✄
(5)
If Equations (1) and (2) are written as a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0,
then we have
a 1 = 5, b 1 = 3, c 1 = –35, a 2 = 2, b 2 = 4, c 2 = –28.Then Equation (5) can be written as x =^1221
12 21
bc b c
ab a b☎
☎
,
Similarly, you can get y =^1221
12 21
ca c a
ab a b☎
☎
By simplyfing Equation (5), we get
x =84 140
20 6
☎ ✆
☎
= 4
Similarly, y =
( 35)(2) (5)( 28)
20 6
✝ ✝ ✝
✝
=
70 140
14
✞ ✟
= 5
Therefore, x = 4, y = 5 is the solution of the given pair of equations.
Then, the cost of an orange is Rs 4 and that of an apple is Rs 5.
Verification : Cost of 5 oranges + Cost of 3 apples = Rs 20 + Rs 15 = Rs 35. Cost of
2 oranges + Cost of 4 apples = Rs 8 + Rs 20 = Rs 28.
Let us now see how this method works for any pair of linear equations in two
variables of the form
a 1 x + b 1 y + c 1 = 0 (1)
and a 2 x + b 2 y + c 2 = 0 (2)
To obtain the values of x and y as shown above, we follow the following steps:
Step 1 : Multiply Equation (1) by b 2 and Equation (2) by b 1 , to get
b 2 a 1 x + b 2 b 1 y + b 2 c 1 = 0 (3)
b 1 a 2 x + b 1 b 2 y + b 1 c 2 = 0 (4)Step 2 : Subtracting Equation (4) from (3), we get:
(b 2 a 1 – b 1 a 2 ) x + (b 2 b 1 – b 1 b 2 ) y + (b 2 c 1 – b 1 c 2 ) = 0