Sum versus reciprocal sum

Let $x,y$ denote two variables, satisfying $\begin{array}{rl} x + y &= 15\\ \dfrac{1}{x} + \dfrac{1}{y} &= \dfrac{3}{10} \end{array}$ where $x,y\neq 0$ . For the second equation, multiply both sides by $10xy$ , so $\begin{array}{rl} 10xy\left(\dfrac{1}{x} + \dfrac{1}{y}\right) &= 10xy \cdot \dfrac{3}{10}\\ 10\left(y + x\right) &= 3xy\\ 10(15) &= 3xy\\ xy &= 50 \end{array}$ which yields $\begin{array}{rl} x + y &= 15\\ xy &= 50 \end{array}$ Relating the system of equations to Vieta's formula (see Note ), we see that the equation to solve is $z^2 - 15z + 50 = 0$ where $x$ and $y$ are the solutions of the equation. Thus, $\begin{array}{rl} (z - 10)(z - 5) &= 0\\ z &= \boxed{5, 10} \end{array}$ which solve the system of equations.

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Note

Vieta's formula for quadratics is $f(x) = {\color{#D61F06}a}z^2 + {\color{#20A900}b}z + {\color{#3D99F6}c}$ For $\begin{array}{rl} x + y &= 15\\ xy &= 50 \end{array}$ we want $-\dfrac{{\color{#20A900}b}}{{\color{#D61F06}a}} = 15$ and $\dfrac{{\color{#3D99F6}c}}{{\color{#D61F06}a}} = 50$ . Clearly, if $\color{#D61F06}a = 1$ , then ${\color{#20A900}b = -15}, {\color{#3D99F6}c = 50}$ .

A Simple, Elegant Solution :

Let $x$ and $y$ represent the two unknown numbers.

Based on the question, we can form a pair of simultaneous equations:

$1$ . $x+y=15$

$2$ . $\frac{1}{x}+ \frac{1}{y}=\frac{3}{10}$

First, we will try to make the second equation a bit more neater and simpler to work with:

$\frac{1}{x}+ \frac{1}{y}=\frac{3}{10}$

$\frac{y+x}{xy}= \frac{3}{10}$

Subsitute value:

$\frac{15}{xy}= \frac{3}{10}$

Do side changing to obtain:

$3xy=10 \times 15$

And you get...

$xy=10 \times 5$

Wallah! We just solved for $x$ and $y$ . If you don't get it...

$x \times y=10 \times 5$

which essentially suggests that the pair of numbers is $10$ and $5$ . However, we can't tell which is which.

And this leads us to the conclusion, that the largest number value is therefore $\boxed{10}$ .