Solve the following pair of linear …

The given pair of the equation is
{tex}\frac{4}{x} + 3y = 14{/tex} ....(1)
{tex}\frac{3}{x} - 4y = 23{/tex} ....(2)
Put {tex}\frac{1}{x} = x{/tex} ....(3)
Then equation (1) and (2) can be rewritten as
4x + 3y = 14 ...(4)
3x - 4y = 23 ...(5)
From equation (5),
4y = 3x - 23
{tex} \Rightarrow \;y = \frac{{3x - 23}}{4}{/tex} ....(6)
Substituting this value of y in equation (4) we get
{tex}4x + 3\left( {\frac{{3x - 23}}{4}} \right) = 14{/tex}
{tex} \Rightarrow{/tex} 25x = 56 + 69 = 125
{tex}\Rightarrow \;x = \frac{{125}}{{25}} = 5{/tex} ....(7)
Substituting this value of x in equation (6), we get
{tex}y = \frac{{3(5) - 23}}{4} = \frac{{15 - 23}}{4}{/tex}
{tex}= \frac{{ - 8}}{4} = - 2{/tex} ...(8)
From equation (3) and equation (7), we get {tex}\frac{1}{x} = 5{/tex}
{tex}\Rightarrow \;\frac{1}{x} = 5{/tex}
Hence, the solution of the given pair of the equation is,{tex}\frac{1}{x} = 5{/tex} y = -2.
Verification. Substituting {tex}\frac{1}{x} = 5{/tex}, y = -2,
We find that both the equations (1) and (2) are satisfied as shown below:
{tex}\frac{4}{x} + 3y = \frac{4}{{\left( {\frac{1}{5}} \right)}} + 3( - 2) = 20 - 6 = 14{/tex}
{tex}\frac{3}{x} - 3y = \frac{3}{{\left( {\frac{1}{5}} \right)}} - 4( - 2) = 15 + 8 = 23{/tex}
Hence, the solution is correct.