Example: Solve the pair of equations: 
	$$\frac{2}{\mathit{x}}+\frac{3}{\mathit{y}}=13$$
	$$\frac{5}{\mathit{x}}-\frac{4}{\mathit{y}}=-2$$
Solution : Write the given pair of equations as
    $$2\left(\frac{1}{\mathit{x}}\right)+3\left(\frac{1}{\mathit{y}}\right)=13$$
	$$5\left(\frac{1}{\mathit{x}}\right)-4\left(\frac{1}{\mathit{y}}\right)=-2$$
These equations are not in the form ax + by + c = 0.
However, if we substitute $\frac{1}{\mathit{x}}=\mathit{p}$ and $\frac{1}{\mathit{y}}=\mathit{q}$;
we get ;     
			$$2\mathit{p}+3\mathit{q}=13$$
			$$5\mathit{p}-4\mathit{q}=-2$$ 

Since $\frac{1}{\mathit{x}}=\mathit{p}$ and $\frac{1}{\mathit{y}}=\mathit{q}$; Substitute the values of p and q 

	$\frac{1}{\mathit{x}}=2$ and $\frac{1}{\mathit{y}}=3$
	Therefore $\mathit{x}=\frac{1}{2}$ and $\mathit{y}=\frac{1}{3}$

Example: Solve the following pair of equations by reducing them to a pair of linear equations :
  $$\frac{5}{\left(\mathit{x}-1\right)}+\frac{1}{\left(\mathit{y}-2\right)}=2$$ 
  $$\frac{6}{\left(\mathit{x}-1\right)}-\frac{3}{\left(\mathit{y}-2\right)}=1$$ 

 Solution:
     
  Let $\frac{1}{\left(\mathit{x}-1\right)}=\mathit{p}$ and $\frac{1}{\left(\mathit{y}-2\right)}=\mathit{q}$
  Subsitute it in the above equations:
  $5\mathit{p}+1\mathit{q}=2$                  (3)
  $6\mathit{p}-3\mathit{q}=1$                   (4)
  
  Now, use any method to solve these equations; we get $\mathit{p}=\frac{1}{3}$ and $\mathit{q}=\frac{1}{3}$
  Now, substituting  $\frac{1}{\left(\mathit{x}-1\right)}=\mathit{p}$ , we get $\frac{1}{\left(\mathit{x}-1\right)}=\frac{1}{3}$ ; $\mathit{x}=4$
  Now, substituting  $\frac{1}{\left(\mathit{y}-2\right)}=\mathit{q}$ , we get $\frac{1}{\left(\mathit{y}-2\right)}=\frac{1}{3}$ ; $\mathit{y}=5$
 Hence, $\mathit{x}=4$, $\mathit{y}=5$ is the required solution of the given pair of equations.
 

Solution : Let the speed of the boat in still water be $\mathit{x}$ km/h and speed of the  stream  be  $\mathit{y}$  km/h.
Then  the speed  of  the  boat  downstream = $\left(\mathit{x}+\mathit{y}\right)$ km/h,
and the speed of the boat upstream = $\left(\mathit{x}\mathit{–}\mathit{y}\right)$ km/h

Since $\mathit{S}\mathit{p}\mathit{e}\mathit{e}\mathit{d}=\frac{\left(\mathit{D}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\right)}{\left(\mathit{T}\mathit{i}\mathit{m}\mathit{e}\right)}$. therefore $\mathit{T}\mathit{i}\mathit{m}\mathit{e}=\frac{\left(\mathit{D}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\right)}{\left(\mathit{S}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\right)}$
In First Case:
Condition 1:when the boat goes 30 km upstream, let the time taken, in hour, be ${\mathit{t}}_1$ . Then 
 ${\mathit{t}}_1=\frac{30}{\left(\mathit{x}-\mathit{y}\right)}$
 
Condition 2:Let  ${\mathit{t}}_2$  be the time, in hours, taken by the boat to go 44 km downstream, then
 ${\mathit{t}}_2=\frac{44}{\left(\mathit{x}+\mathit{y}\right)}$
 
Condition 3: A boat goes 30 km upstream and 44 km downstream in 10  hours, Hence
     ${\mathit{t}}_1+{\mathit{t}}_2=10$
	 $\frac{30}{\left(\mathit{x}-\mathit{y}\right)}+\frac{44}{\left(\mathit{x}+\mathit{y}\right)}=10$                 (1)
	 
In Second Case:
Condition 1: Let ${\mathit{t}}_3$ be the time, in hours, taken by the boat to go 40 km upstream.
     ${\mathit{t}}_3=\frac{40}{\left(\mathit{x}-\mathit{y}\right)}$
Condition 2: Let ${\mathit{t}}_4$ be the time, in hours, taken by the boat to go 55 km downstream.
     ${\mathit{t}}_4=\frac{55}{\left(\mathit{x}+\mathit{y}\right)}$	 
Condition 3: In 13  hours,  it  can  go 40 km upstream   and   55   km down-stream.
		${\mathit{t}}_3+{\mathit{t}}_4=13$
		
		 $\frac{40}{\left(\mathit{x}-\mathit{y}\right)}+\frac{55}{\left(\mathit{x}+\mathit{y}\right)}=13$              (2)
		 
Let $\frac{1}{\left(\mathit{x}-\mathit{y}\right)}=\mathit{u}$ and $\frac{1}{\left(\mathit{x}+\mathit{y}\right)}=\mathit{v}$,              (3)
Subsituting these in equations (1) and (2), pair of linear equations becomes:
$30\mathit{u}+44\mathit{v}=10$                             (4)
$40\mathit{u}+55\mathit{v}=13$                                (5)

Using any of the methods, we get;
$\mathit{u}=\frac{1}{5}$ and $\mathit{v}=\frac{1}{11}$

Putting these values in equation (3);
$\frac{1}{\left(\mathit{x}-\mathit{y}\right)}=\frac{1}{5}$ and $\frac{1}{\left(\mathit{x}+\mathit{y}\right)}=\frac{1}{11}$

i.e.,   $\mathit{x}\mathit{–}\mathit{y}=5$  and  $\mathit{x}+\mathit{y}=11$       (6)  

Solving these equations:
$\mathit{x}=8$ and $\mathit{y}=3$  

Answer:Hence, the speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h.