Solve using elimination.

To solve the system of equations using elimination, we'll start with the given equations:

1. \( 7x + 3y = -16 \) (Equation 1)
2. \( 10x + 8y = -8 \) (Equation 2)

First, we'll manipulate the equations to make elimination easier. We'll aim to eliminate \(y\) by making the coefficients of \(y\) in both equations the same. We can do this by finding a common multiple of the coefficients of \(y\) in both equations (which are 3 and 8).

The least common multiple of 3 and 8 is 24. To achieve this, we'll multiply Equation 1 by 8 and Equation 2 by 3:

Multiply Equation 1 by 8: \[ 8(7x + 3y) = 8(-16) \] \[ 56x + 24y = -128 \quad \text{(Equation 3)} \]

Multiply Equation 2 by 3: \[ 3(10x + 8y) = 3(-8) \] \[ 30x + 24y = -24 \quad \text{(Equation 4)} \]

Now we have the two new equations:

1. \(56x + 24y = -128\) (Equation 3)
2. \(30x + 24y = -24\) (Equation 4)

Next, we subtract Equation 4 from Equation 3 to eliminate \(y\):

\[ (56x + 24y) - (30x + 24y) = -128 - (-24) \] This simplifies to: \[ (56x - 30x) + (24y - 24y) = -128 + 24 \] \[ 26x = -104 \]

Now, we solve for \(x\): \[ x = \frac{-104}{26} = -4 \]

Now that we have \(x\), we can substitute \(x = -4\) back into one of the original equations to solve for \(y\). We'll use Equation 1: \[ 7(-4) + 3y = -16 \] \[ -28 + 3y = -16 \] Add 28 to both sides: \[ 3y = -16 + 28 \] \[ 3y = 12 \] Now divide by 3: \[ y = \frac{12}{3} = 4 \]

Thus, the solution to the system of equations is: \[ \boxed{(-4, 4)} \]