Maths

58 MATHEMATICS
Therefore, x =
– 28)✁
(4)(–35) ✂ (3)( ✂
✂ (5)(4) (3)(2)
i.e., x =
(3)(– 28) (4)( ✄ 35) ✄
✄ (5)(4) (2)(3)
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 =
bc
ab
b ☎ c
,
☎ ab
1 2 2 1
1 2 2 1
☎
☎
Similarly, you can get y =
ca
1 2 ca
2 1
ab ab
By simplyfing Equation (5), we get
Similarly, y =
x =
1 2 2 1
84 ☎ ✆ 140
= 4
☎ 20 6
( 35)(2) (5)( ✝ 28) ✝ ✝
=
✝ 20 6
70 140 ✞ ✟
= 5
14
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
(5)