To solve the system of linear equations:

1. \(2x + 3y = 13\)
2. \(5x + 2y = 27\)

we can use either the substitution method or the elimination method. Here, I'll use the elimination method.

First, let's align the equations for clarity:

\[
\begin{align*}
1. & \quad 2x + 3y = 13 \\
2. & \quad 5x + 2y = 27
\end{align*}
\]

To eliminate one of the variables, we can multiply each equation by a suitable number so that the coefficients of one of the variables are the same (or opposites).

Let's eliminate \(y\). To do this, we can multiply the first equation by 2 and the second equation by 3:

\[
\begin{align*}
1. & \quad 2(2x + 3y) = 2(13) \quad \Rightarrow \quad 4x + 6y = 26 \\
2. & \quad 3(5x + 2y) = 3(27) \quad \Rightarrow \quad 15x + 6y = 81
\end{align*}
\]

Now we have:

\[
\begin{align*}
1. & \quad 4x + 6y = 26 \\
2. & \quad 15x + 6y = 81
\end{align*}
\]

Next, subtract the first equation from the second equation to eliminate \(y\):

\[
(15x + 6y) - (4x + 6y) = 81 - 26
\]

This simplifies to:

\[
11x = 55
\]

Solving for \(x\):

\[
x = \frac{55}{11} = 5
\]

Now that we have \(x = 5\), we can substitute this value back into one of the original equations to solve for \(y\). Let's use the first equation:

\[
2x + 3y = 13
\]

Substitute \(x = 5\):

\[
2(5) + 3y = 13 \quad \Rightarrow \quad 10 + 3y = 13
\]

Solving for \(y\):

\[
3y = 13 - 10 \quad \Rightarrow \quad 3y = 3 \quad \Rightarrow \quad y = \frac{3}{3} = 1
\]

So, the solution to the system of equations is:

\[
x = 5, \quad y = 1
\]

Therefore, the solution is \((5, 1)\).