To solve for \(x\) and \(y\) in the given system of simultaneous equations:

Equation 1: \(-2x - 5y = 3\)

Equation 2: \(x + 3y = 0\)

We can solve this system of equations using the method of substitution or elimination. Let's use the elimination method:

Multiply Equation 2 by 2 to make the coefficients of \(x\) in both equations cancel each other out:

\(2(x + 3y) = 2 \cdot 0\)

\(2x + 6y = 0\)

Now add this modified Equation 2 to Equation 1 to eliminate \(x\):

\((-2x - 5y) + (2x + 6y) = 3 + 0\)

\(-2x + 2x + y = 3\)

\(y = 3\)

Now that we have the value of \(y\), we can substitute it into Equation 2 to solve for \(x\):

\(x + 3(3) = 0\)

\(x + 9 = 0\)

\(x = -9\)

So, the solution for the simultaneous equations is \(x = -9\) and \(y = 3\).