Suppose $f(x,y)$ is a real-valued function such that $f(3x+2y,2x-5y)=19x$, for all real numbers x and y. The value of x for which $f(x,2x)=27$, is

The correct answer is 3.

To solve this problem, we are given the functional equation:
$f(3u+2v,2u-5v)=19u$
for all real numbers $u$ and $v$. We use dummy variables $u$ and $v$ to avoid confusion with the variable $x$ in the expression $f(x,2x)$.

We want to find the expression for $f(x,2x)$. To do this, we set the arguments of the function equal to $x$ and $2x$ respectively:
1) $3u+2v=x$
2) $2u-5v=2x$

Now, we solve this system of linear equations for $u$ in terms of $x$. From equation (1), we can express $v$ as:
$v=\frac{x-3u}{2}$

Substituting this expression for $v$ into equation (2) gives:
$2u-5\left(\frac{x-3u}{2}\right)=2x$

To clear the fraction, we multiply the entire equation by 2:
$4u-5(x-3u)=4x$

Expanding and simplifying the equation:
$4u-5x+15u=4x$
$19u-5x=4x$
$19u=9x$
$u=\frac{9x}{19}$

Since the functional relation is $f(3u+2v,2u-5v)=19u$, substituting the arguments $x$ and $2x$ gives:
$f(x,2x)=19u$

Substitute the value of $u$ in terms of $x$:
$f(x,2x)=19\left(\frac{9x}{19}\right)=9x$

We are given that $f(x,2x)=27$. Therefore:
$9x=27$
$x=3$