Certainly, let's solve the simultaneous equations step by step:

Given equations:

1. \(3x + y = 8\)

2. \(x^2 + xy = 6\)

Let's solve the first equation for \(y\):

\[y = 8 - 3x\]

Substitute this expression for \(y\) into the second equation:

\[x^2 + x(8 - 3x) = 6\]

Simplify the equation:

\[x^2 + 8x - 3x^2 = 6\]

Combine the \(x^2\) terms and the \(x\) terms:

\[-2x^2 + 8x = 6\]

Divide both sides by -2:

\[x^2 - 4x = -3\]

Rearrange the equation:

\[x^2 - 4x + 3 = 0\]

Factor the quadratic equation:

\[(x - 3)(x - 1) = 0\]

This gives us two possible values for \(x\):

1. \(x - 3 = 0 \Rightarrow x = 3\)

2. \(x - 1 = 0 \Rightarrow x = 1\)

Now, substitute these values of \(x\) back into the equation \(y = 8 - 3x\) to find the corresponding values of \(y\):

For \(x = 3\): \(y = 8 - 3(3) = -1\)

For \(x = 1\): \(y = 8 - 3(1) = 5\)

Thus, the two pairs of values that satisfy the simultaneous equations are:

1. \(x = 3\) and \(y = -1\)

2. \(x = 1\) and \(y = 5\)

The correct answer is -1 and 5