# 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 $y$?

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Replace $x$ with $\frac {10}y$ in $x=2y+1$. then multiply by $y$ on both sides of the resulting equality. – Git Gud Oct 12 '13 at 22:15
@GitGud What i get is (10/y) = 2y+1, then i multiply by y and i get 10 = 2y^2+1, and i doing something wrong? – Orel Oct 12 '13 at 22:19
@Orel No, that's what you're supposed to do, but it's $10=2y^2+y$ – egreg Oct 12 '13 at 22:20
$10=2y^2+y$. Write this as $2y^2+y-10=0$. You have a quadratic equation here... – David Mitra Oct 12 '13 at 22:20

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 $0$.

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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.

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Isn't substituting $2y+1$ for $x$ in $xy=10$ easier? It's just the same, but with less computations. – egreg Oct 12 '13 at 22:24
Yes, it is. Even i had same thought but i went through the way he\she started. – Shravan40 Oct 12 '13 at 22:29

$$1 = \left(x - 2y\right)^{2} = x^{2} - 4xy + 4y^{2}$$

$$1 + 80 = x^{2} + 4xy + 4y^{2} = \left(x + 2y\right)^{2}$$

$$x + 2y = \pm 9\,, \quad x = {1 \pm 9 \over 2}\,, \quad y = {\pm 9 - 1 \over 4}$$

$$\color{#ff0000}{\large\left(x, y\right) = \quad \left(5,2\right)\,,\quad \left(-4, -\,{5 \over 2}\right)}$$

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