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