Solve the following subtraction problems with checks using addition: c) 75 - 68 d) 90 - 47

$$5 - 8$$

Since $5 < 8$, borrow 1 ten from the tens place. The 7 in the tens place becomes 6, and the 5 in the ones place becomes 15.

$$15 - 8 = 7$$

$$6 - 6 = 0$$

The result of the subtraction is 7.

$$\begin{array}{c} \text{T O} \\ \ 7 \ 5 \\ - \ 6 \ 8 \\ \hline \ \mathbf{0 \ 7} \\ \end{array}$$

Check (Addition)

$$7 + 68$$

Add the ones place: $7 + 8 = 15$. Write down 5 and carry over 1 to the tens place.
Add the tens place: $1 + 0 + 6 = 7$.
The sum is 75, which matches the original number.

$$\begin{array}{c} \text{T O} \\ \ 0 \ 7 \\ + \ 6 \ 8 \\ \hline \ \mathbf{7 \ 5} \\ \end{array}$$

$$0 - 7$$

Since $0 < 7$, borrow 1 ten from the tens place. The 9 in the tens place becomes 8, and the 0 in the ones place becomes 10.

$$10 - 7 = 3$$

$$8 - 4 = 4$$

The result of the subtraction is 43.

$$\begin{array}{c} \text{T O} \\ \ 9 \ 0 \\ - \ 4 \ 7 \\ \hline \ \mathbf{4 \ 3} \\ \end{array}$$

Check (Addition)

$$43 + 47$$

Add the ones place: $3 + 7 = 10$. Write down 0 and carry over 1 to the tens place.
Add the tens place: $1 + 4 + 4 = 9$.
The sum is 90, which matches the original number.

$$\begin{array}{c} \text{T O} \\ \ 4 \ 3 \\ + \ 4 \ 7 \\ \hline \ \mathbf{9 \ 0} \\ \end{array}$$

Complete the subtraction and addition facts

$$78 - 65$$

Subtract the ones place: $8 - 5 = 3$.
Subtract the tens place: $7 - 6 = 1$.

$$78 - 65 = \boxed{\mathbf{13}}$$

The result of the subtraction plus the number subtracted should equal the original number.

$$\boxed{\mathbf{13}} + 65 = 78$$

$$88 - 59$$

Subtract the ones place: $8 - 9$. Since $8 < 9$, borrow 1 ten from the tens place. The 8 in the tens place becomes 7, and the 8 in the ones place becomes 18.

$$18 - 9 = 9$$

Subtract the tens place: $7 - 5 = 2$.

$$88 - 59 = \boxed{\mathbf{29}}$$

The result of the subtraction plus the number subtracted should equal the original number.

$$\boxed{\mathbf{29}} + 59 = 88$$