Let,
the numbers be "x" and "y"
According to the question,
x = 4y ......................................................equation (1)
\(\frac{1}{x} + \frac{1}{y}= \frac{45}{4}\)...................................equation (2)
Taking equation (2)
\(\frac{1}{x} + \frac{1}{y}= \frac{45}{4}\)
Substituting the value of "x" from equation (1), we get,
\(\frac{1}{(4y)} + \frac{1}{y}= \frac{45}{4}\)
\(\frac{1}{4y} + \frac{1}{y}* \frac{4}{4} = \frac{45}{4}\)
\(\frac{1}{4y} + \frac{4}{4y}= \frac{45}{4}\)
\(\frac{1+4}{4y} = \frac{45}{4}\)
\(\frac{5}{4y} = \frac{45}{4}\)
Cross multiplying, we get,
\(5*4 = 45*4y\)
\(20 = 180y\)
\(\frac{20}{180} = y\)
\(\frac{1}{9} = y\)
\(y = \frac{1}{9}\)
Now,
Taking equation (1)
x = 4y
Substituting the value of "y", we get
\(x = 4( \frac{1}{9} )\)
\(x= \frac{4}{9}\)
So, the numbers are \(\frac{1}{9}\) and \(\frac{4}{9}\)
