(iii)
1/3x + 1/5y = 1/15; 1/2x + 1/3y = 1/12
1/3x
+ 1/5y = 1/15
Multiplying
the equation by 15 on both the sides we get,
5/x
+ 3/y = 1
1/2x
+ 1/3y = 1/12
Multiplying
the equation by 12 on both the sides we get,
6/x
+ 4/y = 1
Now,
Let 1/x = a and 1/y = b
∴ We get, 5a + 3b =
1 ......... eq. no. (1)
and 6a + 4b = 1
........ eq. no. (2)
Multiplying
(1) bt 4, we get
20a
+ 12b = 4 .........(3)
Multiplying
(2) by 3, we get
18a
+ 12b = 3 ........(vi)
Subtracting (4) from (3),
|
20a
|
+12b
|
=
|
4
|
|
18a
|
+12b
|
=
|
3
|
|
(-)
|
(-)
|
|
(-)
|
|
2a
|
|
=
|
1
|
∴ a = 1/2
Substituting
a = 1/2 in equation (2)
∴ 6a + 4b = 1
∴ 6(1/2) + 4b = 1
∴ 3 + 4b = 1
∴ 4b = 1 – 3
∴ 4b = - 2
∴ b= - 2 /4
∴ b = -1/2
Substituting
the values of a and b,
|
a = 1/2
∴ 1/x = 1/2
∴ x = 2
|
|
|
b = -1/2
∴ 1/y = -1/2
∴ y = -2
|
∴ x = 2 and y = - 2 is
the solution of given simultaneous equations.
