Solution:

Let us write the given pair of equations as

`2 (1/x) + 3 ( 1/y) = 13` (1)

`5 (1/x) -4 ( 1/y) = -2` (2)

These equations are not in the form `ax + by + c = 0`. However, if we substitute

`1/x = p` and `1/y = q` in Equations (1) and (2), we get

`2p + 3q = 13` (3)

`5p – 4q = – 2` (4)

So, we have expressed the equations as a pair of linear equations. Now, you can use
any method to solve these equations, and get `p = 2, q = 3`.

You know that `p = 1/x` and `q = 1/y` .

Substitute the values of p and q to get

`1/x = 2` , i.e., `x = 1/2` and `1/y =3` , i.e., ` y = 1/3` .

Verification : By substituting `x = 1/2 ` and `y = 1/3` in the given equations, we find that
both the equations are satisfied.

---

Solution:

Let us put ` 1/(x-1) = p` and `1/(y-2) = q` Then the given equations

` 5(1/(x-1) ) + 1/(y-2) =2` (1)

`6 (1/(x-1) ) -3 (1/(y-2) ) =1` (2 )


can be written as : `5p + q = 2` (3)

`6p – 3q = 1 `(4 )

Equations (3) and (4) form a pair of linear equations in the general form. Now,
you can use any method to solve these equations. We get ` p = 1/3` and `q = 1/3` .

Now, substituting `1/(x-1)` for p, we have

`1/(x-1)= 1/3` ,

i.e., `x – 1 = 3`, i.e., `x = 4`.

Similarly, substituting ` 1/(y-2)` for q, we get

`1/(y-2) = 1/3`

i.e., `3 = y – 2`, i.e.,` y = 5`
Hence, x = 4, y = 5 is the required solution of the given pair of equations.

Verification : Substitute x = 4 and y = 5 in (1) and (2) to check whether they are
satisfied .

---

Solution:

Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Then the
speed of the boat downstream

= (x + y) km/h,


and the speed of the boat upstream = (x – y) km/h

Also, ` text (time) = (text ( distance ) )/( text ( speed ) )`

In the first case, when the boat goes 30 km upstream, let the time taken, in hour,
be `t_1`. Then

`t_1 = 30/(x-y)`

Let `t_2` be the time, in hours, taken by the boat to go 44 km downstream. Then

`t_2 = 44/(x+y)` . The total time taken, `t_1 + t_2`, is 10 hours. Therefore, we get the equation

`30/(x-y) + 44/(x+y) = 10` (1)

In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. We
get the equation

`40/(x-y) + 55/(x+y) = 13` (2)

Put `1/(x-y) = u` and ` 1/(x+ y ) = v` (3)

On substituting these values in Equations (1) and (2), we get the pair of linear
equations:

`30u + 44v = 10` or `30u + 44v – 10 = 0` (4)

`40u + 55v = 13` or `40u + 55v – 13 = 0` (5)

Using Cross-multiplication method, we get

`u/(44 (-13) -55 (-10 )) = v/(40 (-10) -30 ( -13 ) ) = 1/(30 (55) -44 (40 ) )`

i.e., `u/(-22) = v/(-10) = 1/(-110)`

i.e. ` u = 1/5 , v = 1/11`

Now put these values of u and v in Equations (3), we get

`1/(x-y) = 1/5` and ` 1/(x+y) = 1/11`

i.e., ` x-y = 5` and `x+ y = 11` (6)


Adding these equations, we get
2x = 16
i.e., x = 8
Subtracting the equations in (6), we get
2y = 6
i.e., y = 3

Hence, the speed of the boat in still water is 8 km/h and the speed of the stream
is 3 km/h.
Verification : Verify that the solution satisfies the conditions of the problem.

---