(ii)
2/x + 6/y = 13; 3/x + 4/y = 12
Let 1/x = a and 1/y = b
∴ We get, 2a + 6b =
13 ......... eq. no. (1)
and 3a + 4b = 12
........ eq. no. (2)
Multiplying
(1) by 2,
4a
+ 12b= 26 .......(3)
Multiplying
(2) by 3,
9a
+ 12b= 36 ......(4)
Subtracting
(3) from (4),
|
9a
|
+12b
|
=
|
36
|
|
4a
|
+12b
|
=
|
26
|
|
(-)
|
(-)
|
|
(-)
|
|
5a
|
|
=
|
10
|
∴ a = 10/5
∴ a = 2
Substituting
a = 2 in (1),
2
(2) + 6b = 13
∴ 4 + 6b = 13
∴ 6b = 13 – 4
∴ b = 9/6
∴ b = 3/2
Substituting
the values of a and b,
|
a = 2
∴ 1/x = 2
∴ x = 1/2
|
|
|
b = 3/2
∴ 1/y = 3/2
∴ y = 2/3
|
∴ x = 1/2 and y = 2/3 is
the solution of given simultaneous equations.