To solve the system of equations:

1. \( 2x = 5y \) (Equation 1)
2. \( x^2 + y^2 = 261 \) (Equation 2)

First, we can express \( x \) in terms of \( y \) from Equation 1:

\[
x = \frac{5y}{2}
\]

Next, we substitute \( x \) in Equation 2:

\[
\left(\frac{5y}{2}\right)^2 + y^2 = 261
\]

Expanding this, we have:

\[
\frac{25y^2}{4} + y^2 = 261
\]

To combine the terms, we need a common denominator:

\[
\frac{25y^2}{4} + \frac{4y^2}{4} = 261
\]
\[
\frac{29y^2}{4} = 261
\]

Now, multiply both sides by 4:

\[
29y^2 = 1044
\]

Next, divide by 29:

\[
y^2 = \frac{1044}{29} \approx 36
\]

Taking the square root, we find:

\[
y = 6 \quad \text{or} \quad y = -6
\]

Now we substitute \( y \) back into Equation 1 to find \( x \):

1. If \( y = 6 \):

\[
2x = 5(6) \quad \Rightarrow \quad 2x = 30 \quad \Rightarrow \quad x = 15
\]

2. If \( y = -6 \):

\[
2x = 5(-6) \quad \Rightarrow \quad 2x = -30 \quad \Rightarrow \quad x = -15
\]

Thus, the solutions are:

\[
(x, y) = (15, 6) \quad \text{and} \quad (x, y) = (-15, -6)
\]