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The sum of two numbers is $$11$$ and the sum of their reciprocals is $$\dfrac{{11}}{{28}}$$ Find the numbers.
Given sum of two numbers is 11
x+y=11 _____ (1)
sum of their reciprocals is $$\frac{11}{28}$$.
$$\frac{1}{x}+\frac{1}{y}=\frac{11}{48}$$
$$\frac{y+x}{xy}=\frac{11}{48}$$
$$28(x+y)= 11xy$$ _______ (2)
$$28(11)= 11.xy$$
$$xy=28$$
Now
$$(x-y)^{2}=x^{2}+y^{2}-2xy$$
$$(x-y)^{2}=(x+y)^{2}-4xy$$
$$(x-y)^{2}=(11)^{2}-4.28$$
$$(x-y)^{2}=121-112$$
$$(x-y)^{2}=9$$
$$x-y=\sqrt{9}$$
$$x-y=3$$ _______ (3)
On adding (1) & (3)
x+y=1
x-y=3
_________
2x=14
x=14/2
x=7
putting x=7 in equation _____ (1)
x+y=11
$$\pi +y=11$$
$$y=11-7$$
$$y=4$$
$$\therefore $$ The two numbers are $$(x,y)=(7,4)$$