Write a system of equations to describe the situation below,solve using any method, and fill in the blanks. Darrell and Zoe decided to shoot arrows at a simple target with a large outer ring and a smaller bull's-eye Darrell went first and landed 5 arrows in the outer ring and 4 arrows in the bull's-eye, for a total of 265 points. Zoe went second and got 5 arrows in the outer ring and 2 arrows in the bull's -eye, earning a total of 165 points How many points is each region of the target worth? The outer ring is worth square points, and the bull's-eye is worth square points.

Let's denote the points for the outer ring as $$x$$ and the points for the bull's-eye as $$y$$ . We can set up the following system of equations based on the information provided:

$$\begin{align*} 5x + 4y &= 265 \quad \text{(Darrell's score)} \\ 5x + 2y &= 165 \quad \text{(Zoe's score)} \end{align*}$$

To solve this system, we can use the elimination method. We can subtract the second equation from the first equation to eliminate $$x$$ :

$$(5x + 4y) - (5x + 2y) = 265 - 165$$

Simplifying the equation, we get:

$$2y = 100$$

Dividing both sides by 2, we find:

$$y = 50$$

Now that we have the value of $$y$$ , we can substitute it back into one of the original equations to solve for $$x$$ . Let's use the second equation:

$$5x + 2(50) = 165$$

Simplifying the equation, we get:

$$5x + 100 = 165$$

Subtracting 100 from both sides, we have:

$$5x = 65$$

Dividing both sides by 5, we find:

$$x = 13$$

Therefore, the outer ring is worth $$\boxed{13}$$ points, and the bull's-eye is worth $$\boxed{50}$$ points.