Question

classify the sum: 4z/z + z-3/6
rational number
irrational number
whole number
integer

Answer 1

To classify the expression \( \frac{4z}{z} + \frac{z - 3}{6} \), we first simplify it.

1. Simplify \( \frac{4z}{z} \):
- This simplifies to \( 4 \) (as long as \( z \neq 0 \)).

2. The second term \( \frac{z - 3}{6} \) cannot be simplified further without knowing the value of \( z \).

Combining these two terms results in:
\[
4 + \frac{z - 3}{6}
\]

Next, let's express the entire sum in a single fraction:
\[
4 + \frac{z - 3}{6} = \frac{24}{6} + \frac{z - 3}{6} = \frac{24 + (z - 3)}{6} = \frac{z + 21}{6}
\]

Now, we need to determine the classification based on the value of \( z \):

- If \( z \) is a rational number, then \( z + 21 \) is a rational number, and thus \( \frac{z + 21}{6} \) is also a rational number.
- If \( z \) is an integer, \( z + 21 \) will still be an integer, making \( \frac{z + 21}{6} \) a rational number, which may or may not be an integer based on the value of \( z \).
- If \( z \) is a whole number or zero, it still remains rational.
- If \( z \) is an irrational number, then \( z + 21 \) is irrational, and \( \frac{z + 21}{6} \) becomes irrational.

### Conclusion:
The sum \( \frac{4z}{z} + \frac{z - 3}{6} \) can be classified as a rational number if \( z \) is rational, or an irrational number if \( z \) is irrational.

Thus, the final classification will depend on the value of \( z \). Without specific information about \( z \):

- If \( z \) is rational: **Rational number**
- If \( z \) is irrational: **Irrational number**

It cannot be classified strictly as a whole number or an integer without knowing more about \( z \).