Solve these subtraction problems with borrowing: a) 968 - 589, b) 802 - 279

We cannot subtract 9 from 8, so we borrow 1 from the tens place. The 6 becomes 5, and the 8 becomes 18. $18 - 9 = 9$

We cannot subtract 8 from the remaining 5, so we borrow 1 from the hundreds place. The 9 becomes 8, and the 5 becomes 15. $15 - 8 = 7$

$$
8 - 5 = 3
$$
Combining these, we get:
$$
\begin{array}{c}
\phantom{0}\text{8}\phantom{0}\text{15}\phantom{0}\text{18} \\
\cancel{9}\cancel{6}\cancel{8} \\
-589 \\
\hline
\boxed{\text{379}}
\end{array}
$$

2. Subtract 279 from 802.

We cannot subtract 9 from 2, and we cannot borrow from the tens place (which is 0). So, we borrow 1 from the hundreds place. The 8 becomes 7, and the 0 in the tens place becomes 10.
Now, we borrow 1 from the tens place. The 10 becomes 9, and the 2 in the ones place becomes 12. $12 - 9 = 3$

The tens digit is now 9. $9 - 7 = 2$

The hundreds digit is now 7. $7 - 2 = 5$ Combining these, we get: $\begin{array}{c} \phantom{0}\text{7}\phantom{0}\text{9}\phantom{0}\text{12} \\ \cancel{8}\cancel{0}\cancel{2} \\ -279 \\ \hline \boxed{\text{523}} \end{array}$