how can I solve this kind of equation? I've got a system of equations which is:
$\begin{cases} x=2y+1\\xy=10\end{cases}$
I have gone into this: $x=\dfrac {10}y$.

How can I find the $x$ and $y$?
 A: Hint : 
This kind of equation can be solved by substituting the value of $ x $ or $ y $ in the first equation.And the above equation will become quadratic, solve for it
$ x = 2y +1 \dots (1)$
$xy = 10 $
$ \implies x = \frac{10}{y}$
Put the value of x in equation (1)
$ \frac{10}{y} = 2y+1 $
$ 10 = 2y^2 + y $
$ 2y^2 + y -10 = 0 \dots(2)$
Solve this quadratic equation, For each value of $y$ you will get a $x$
Same you can do it by replacing $ y = \frac{10}{x}$
Hope, you can proceed from here. 
A: Notice that $10 = xy = (2y + 1)y = 2y^2 + y$. But then $$2y^2 + y - 10 = 0.$$
Can you solve this quadratic equation?
If you use the substitutions $x = \frac{10}{y}$ or $y = \frac{10}{x}$ then you are implicitly assuming either $y$ or $x$ is not $0$.
A: $$
1 = \left(x - 2y\right)^{2} = x^{2} - 4xy + 4y^{2}
$$
$$
1 + 80 = x^{2} + 4xy + 4y^{2} = \left(x + 2y\right)^{2}
$$
$$
x + 2y = \pm 9\,,
\quad
x = {1 \pm 9 \over 2}\,,
\quad
y = {\pm 9 - 1 \over 4}
$$
$$
\color{#ff0000}{\large\left(x, y\right)
=
\quad
\left(5,2\right)\,,\quad
\left(-4, -\,{5 \over 2}\right)}
$$
