# How to solve this system of equations?

## Real numbers x and y satisfy the system of equations $x + y + \frac{x}{y} = 10$ $\frac{x}{y} \cdot \left(x + y\right) = 20$ Determine the sum of all possible values of the expression x+y.

Jan 12, 2017

$x = \frac{20 \left(6 + \sqrt{5}\right)}{31} = 5.314$ and
$y = \frac{35 + 11 \sqrt{5}}{31} = 1.922$
or
$x = \frac{20 \left(6 - \sqrt{5}\right)}{31} = 2.428$ and
$y = \frac{35 - 11 \sqrt{5}}{31} = 0.336$

#### Explanation:

In the system of equations $x + y + \frac{x}{y} = 10$ and $\frac{x}{y} \left(x + y\right) = 20$, let us assume $x + y = u$ and $\frac{x}{y} = v$, then the equations become

$u + v = 10$ and $u v = 20$. The latter gives $v = \frac{20}{u}$ and substituting it in former, we get

$u + \frac{20}{u} = 10$ or ${u}^{2} - 10 u + 20 = 0$

and using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ we get

$u = \frac{10 \pm \sqrt{100 - 4 \times 1 \times 20}}{2} = \frac{10 \pm \sqrt{20}}{2} = 5 \pm \sqrt{5}$

As $u + v = 10$, if $u = 5 + \sqrt{5}$, $v = 5 - \sqrt{5}$

and if $u = 5 - \sqrt{5}$, $v = 5 + \sqrt{5}$

(1) In first case we have $x + y = 5 + \sqrt{5}$ and $\frac{x}{y} = 5 - \sqrt{5}$. The latter means $y = \frac{x}{5 - \sqrt{5}}$ and hence

$x + \frac{x}{5 - \sqrt{5}} = 5 + \sqrt{5}$ or $x \left(6 - \sqrt{5}\right) = 20$ or

$x = \frac{20}{6 - \sqrt{5}} = \frac{20 \left(6 + \sqrt{5}\right)}{31} = 5.314$ and

$y = \frac{20 \left(6 + \sqrt{5}\right)}{31 \left(5 - \sqrt{5}\right)} = \frac{20 \left(6 + \sqrt{5}\right) \left(5 + \sqrt{5}\right)}{31 \cdot 20}$

= $\frac{35 + 11 \sqrt{5}}{31} = 1.922$

(2) In other case we have $x + y = 5 - \sqrt{5}$ and $\frac{x}{y} = 5 + \sqrt{5}$. The latter means $y = \frac{x}{5 + \sqrt{5}}$ and hence

$x + \frac{x}{5 + \sqrt{5}} = 5 - \sqrt{5}$ or $x \left(6 + \sqrt{5}\right) = 20$ or

$x = \frac{20}{6 + \sqrt{5}} = \frac{20 \left(6 - \sqrt{5}\right)}{31} = 2.428$ and

$y = \frac{20 \left(6 - \sqrt{5}\right)}{31 \left(5 + \sqrt{5}\right)} = \frac{20 \left(6 - \sqrt{5}\right) \left(5 - \sqrt{5}\right)}{31 \cdot 20}$

= $\frac{35 - 11 \sqrt{5}}{31} = 0.336$
graph{(x+y+x/y-10)(x/y(x+y)-20)=0 [-10, 10, -5, 5]}

Jan 12, 2017

$\left(\begin{matrix}x & y \\ \frac{20}{31} \left(6 + \sqrt{5}\right) & \frac{1}{31} \left(35 + 11 \sqrt{5}\right) \\ \frac{20}{31} \left(6 - \sqrt{5}\right) & \frac{1}{31} \left(35 - 11 \sqrt{5}\right)\end{matrix}\right)$

#### Explanation:

Calling $\frac{x}{y} = a$ and $x + y = b$ we have

$\left\{\begin{matrix}a + b = 10 \\ a b = 20\end{matrix}\right.$

solving we have

$a = 5 - \sqrt{5} , b = 5 + \sqrt{5}$ and
$a = 5 + \sqrt{5} , b = 5 - \sqrt{5}$

or

$\left\{\begin{matrix}\frac{x}{y} = 5 - \sqrt{5} \\ x + y = 5 + \sqrt{5}\end{matrix}\right.$

and

$\left\{\begin{matrix}\frac{x}{y} = 5 + \sqrt{5} \\ x + y = 5 - \sqrt{5}\end{matrix}\right.$

but

$\left\{\begin{matrix}\frac{x}{y} = a \\ x + y = b\end{matrix}\right.$

for $x , y$ gives

$\left\{\begin{matrix}x = \frac{a b}{1 + a} \\ y = b - \frac{a b}{1 + a}\end{matrix}\right.$

so we have finally

$\left(\begin{matrix}x & y \\ \frac{20}{31} \left(6 + \sqrt{5}\right) & \frac{1}{31} \left(35 + 11 \sqrt{5}\right) \\ \frac{20}{31} \left(6 - \sqrt{5}\right) & \frac{1}{31} \left(35 - 11 \sqrt{5}\right)\end{matrix}\right)$