To solve this system of equations, we can use the method of substitution or elimination.

Let's use the method of elimination.

First, we can multiply the first equation by 4 and the second equation by 8 to make the coefficients of x in both equations equal:

(4)(8x + 3y = -21) -> 32x + 12y = -84
(8)(4x + 5y = -7) -> 32x + 40y = -56

Now, we can subtract the first equation from the second equation:

(32x + 40y) - (32x + 12y) = -56 - (-84)
28y = 28
y = 1

Now that we have found the value of y, we can substitute it back into one of the original equations to solve for x. Let's substitute y = 1 into the first equation:

8x + 3(1) = -21
8x + 3 = -21
8x = -24
x = -3

Therefore, the solution to the system of equations is x = -3 and y = 1.