Question

x+1/x=4, y-1/y=√5 then xy+1/xy-√15=?​

Answer 1

Given:

$\frac{x+1}{x}=4\\\\ \frac{y-1}{y}=\sqrt{5}$

To find:

$\frac{xy+1}{xy}-\sqrt{15}=?$

Solution:

Let:

$\frac{x+1}{x}=4...............(i)\\\\ \frac{y-1}{y}=\sqrt{5}............(ii)$

$\frac{xy+1}{xy}-\sqrt{15}=?...............(iii)$

solve equation (i):

$\to \frac{x+1}{x}=4\\\\ \to x+1=4x\\\\ \to 1=4x-x\\\\\to 3x=1\\\\\to x= \frac{1}{3}$

solve equation (ii):

$\to \frac{y-1}{y}=\sqrt{5} \\\\ \to 1 -\frac{1}{y}= \sqrt{5} \\\\ \to 1 -\sqrt{5} =\frac{1}{y} \\\\ \to y =\frac{1}{(1 -\sqrt{5})}$

after solve equation..(iii) put the value of x and y:

$\to 1+ \frac{1}{\frac{1}{3(1-\sqrt{5})}}-\sqrt{15}\\\\\to 1+ \frac{1}{\frac{1}{(3-3\sqrt{5})}}-\sqrt{15}\\\\\to 1+ 3-3\sqrt{5}-\sqrt{15}\\\\\to 4-3\sqrt{5}-\sqrt{15}$

Final answer is: 4-3√5-√15