Conflicting Coefficients: Solve the System of Equations -8x + 5y = 3 and 10x + y = 16

Solve the following system of equations:

$\begin{cases} -8x+5y=3 \\ 10x+y=16 \end{cases}$

To solve this system of equations, we will use the elimination method.

The system of equations is:

$\begin{cases}-8x+5y=3 \\ 10x+y=16 \end{cases}$

We will first make the coefficients of $y$ the same so that we can eliminate $y$. To do that, we need both equations to have the same coefficient for $y$. The first equation already has $5y$, so we will multiply the second equation by 5:

$5(10x + y) = 5 \times 16$

This gives the equation:

$50x + 5y = 80$

Now the system is:

$\begin{cases} -8x + 5y = 3 \\ 50x + 5y = 80 \end{cases}$

We will subtract the first equation from the second to eliminate $y$:

$(50x + 5y) - (-8x + 5y) = 80 - 3$

Solving this, we get:

$50x - (-8x) + 5y - 5y = 80 - 3$

$58x = 77$

Thus, the value of $x$ is:

$x = \frac{77}{58} \approx 1.32$

Now, we substitute this value back into one of the original equations to find $y$. It's often easier to substitute into the simpler equation, $10x + y = 16:$

$10(1.32) + y = 16$

$13.2 + y = 16$

Solving for $y$, we have:

$y = 16 - 13.2 = 2.8$

Therefore, the solution to the system of equations is:

$x = 1.32, y = 2.8$

This corresponds to the given correct answer choice.