On a hypothetical computer with a word length of three digits and truncation, compute the solution of a system of equations 
On a hypothetical computer with a word length of three digits and truncation, compute the solution of
$$
        \begin{matrix}
        -3x & + & y & = & -2 \\
        10x & - & 3y & = & 7 \\
        \end{matrix}
$$
a. Without partial pivoting
b. With partial pivoting
c. Exactly

We consider the matrix $$
        \begin{bmatrix}
        -3 & 1 & -2 \\
        10 & -3 & 7 \\
        \end{bmatrix}
$$
and the operation $R_2 → R_2 – (\frac{a_{kj}}{a_{ij}})R_1:$ where $R_i$ are rows and $k > i, a_{ij} \neq 0.$ Then the new matrix looks like 
$$
        \begin{bmatrix}
        -3 & 1 & -2 \\
        10 – (-\frac {10}{3})(-3)  & -3 & 7 - (-\frac73)(-2) \\
        \end{bmatrix}
$$
We'll have to truncate $\frac {10}{3} = 3.3333333333333335$ to $3.333 = 0.333 \times 10^1$. In the same  way, $\frac 73 = 0. 233 \times 10^1$, so this matrix is 
$$
        \begin{bmatrix}
        -3 & 1 & -2 \\
        10 – (-0.333 \times 10^1)(-3)  & -3 & 7 - (-0. 233 \times 10^1)(-2) \\
        \end{bmatrix}
$$
which is
$$
        \begin{bmatrix}
        -3 & 1 & -2 \\
        0.01  & -3 & 2.34 \\
        \end{bmatrix}
$$
Because of the truncation $0$ becomes $0.01$, so that if we solve  $0.01x - 3y =  2.34$ for $(x, y)$ we get distorted final solution. 
Now consider partial pivoting which calls for row swapping. Then our matrix is 
$$
        \begin{bmatrix}
        10 & -3 & 7  \\
        -3 & 1 & -2 \\
        \end{bmatrix}
$$
Applying the row operation $R_2 → R_2 – (\frac{a_{kj}}{a_{ij}})R_1$ yields 
$$
        \begin{bmatrix}
        10 & -3 & 7 \\
        -3 - (-\frac {3}{10})10   & 1 – (\frac {1}{10})(-3)  & -2 - (-\frac{1}{5})7\\
        \end{bmatrix}
$$
We don't have to truncate $\frac {3}{10}, \frac {1}{10}, \frac{1}{5}$ because they are $0.3, 0.1, 0.2$ respectively. So, the matrix is 
$$
        \begin{bmatrix}
        10 & -3 & 7 \\
        0   & 1.3  & -0.6\\
        \end{bmatrix}
$$
Then $y = \frac {-0.6}{1.3} = -0.4615384615384615$. Since we didn't lose any digits by truncation, this $y$ is also exact. 
Does that make sense?

You have arrived at the correct conclusion. It is however possible to argue that you have made mistakes and reduce your score during a test. This depends on how strict an instructor you have. You specifically state that $\frac{10}{3}$ has a finite decimal expansion and that the last digit is the digit $5$. Of course, you know that this is wrong, but you should have written $\frac{10}{3} = 3.\overline{3} \approx 3.33 \cdot 10^{-1}$. Moreover, it is possible to argue that while you computer makes large errors, it can keep track of the entries which should be zero. In fact this is what is done in all library implementations of Gaussian elimination. They do not compute those entries which should be zero. In fact, those memory locations are used to store the multipliers obtained during the process, i.e. the matrix $A$ is overwritten by its $LU$ factorization.
