Question

The sum of two numbers is $11$ and the sum of their reciprocals is $\frac{11}{28}$ Find the numbers.

Answer 1

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)

$28\left(11\right)=11.xy$

$xy=28$

Now

$(x-y{)}^2=x^2+y^2-2xy$

$(x-y{)}^2=(x+y{)}^2-4xy$

$(x-y{)}^2=(11{)}^2-4\times 28$

$(x-y{)}^2=121-112$

$(x-y{)}^2=9$

$x-y=\sqrt{9}$

$x-y=3$ _______ (3)

On adding (1) & (3)

$x+y=1$

$x-y=3$

_________

$2x=14$

$x=\frac{14}{2}$

$x=7$

putting $x=7$ in equation _____ (1)

$x+y=11$

$7+y=11$

$y=11-7$

$y=4$

$\therefore$ The two numbers are $(x,y)=(7,4)$