Solution for the system defined by the set of equations 4y + 3z = 8; 2x – z = 2 and 3x + 2y = 5 is A. x = 0; y = 1; z = \[\frac{4}{3}\] B. x = 0; y = \[\frac{1}{2}\]; z = 2 C. x = 1; y = \[\frac{1}{2}\]; z = 2 D. non-existent

” option2=”x = 0; y = \[\frac{1}{2}\]; z = 2″ option3=”x = 1; y = \[\frac{1}{2}\]; z = 2″ option4=”non-existent” correct=”option2″]

The correct answer is $\boxed{\text{A}}$.

To solve the system of equations, we can use the elimination method. First, let’s eliminate $z$. We can do this by adding the first and third equations together. This gives us the equation $6y + 3x = 10$.

Next, let’s eliminate $x$. We can do this by subtracting the second equation from the third equation. This gives us the equation $y = \frac{1}{2}$.

Now that we know $y = \frac{1}{2}$, we can substitute this value into the first equation to solve for $x$. This gives us the equation $4 \cdot \frac{1}{2} + 3z = 8$. Solving for $z$, we get $z = \frac{4}{3}$.

Therefore, the solution to the system of equations is $x = 0$, $y = \frac{1}{2}$, and $z = \frac{4}{3}$.

Here is a step-by-step solution:

1. Add the first and third equations together. This gives us the equation $6y + 3x = 10$.
2. Subtract the second equation from the third equation. This gives us the equation $y = \frac{1}{2}$.
3. Substitute $y = \frac{1}{2}$ into the first equation to solve for $x$. This gives us the equation $4 \cdot \frac{1}{2} + 3z = 8$. Solving for $z$, we get $z = \frac{4}{3}$.
4. Therefore, the solution to the system of equations is $x = 0$, $y = \frac{1}{2}$, and $z = \frac{4}{3}$.