Given polynomials \( p \), \( q \), and \( r \) such that \( q \neq 0 \):

[tex]\[
\frac{p}{q} + \frac{r}{q} = \square
\][/tex]

[tex]\[
\frac{p}{q} - \frac{r}{q} = \square
\][/tex]



Answer :

Certainly! Let's solve the given problems step-by-step:

### Problem 1: \(\frac{p}{q} + \frac{r}{q}\)

1. Given: We have the fractions \(\frac{p}{q}\) and \(\frac{r}{q}\) where \(q \neq 0\).

2. Common Denominator: Notice that both fractions have the same denominator \(q\).

3. Addition: When adding fractions with the same denominator, we simply add the numerators together while keeping the denominator the same.
[tex]\[ \frac{p}{q} + \frac{r}{q} = \frac{p + r}{q} \][/tex]

Thus,
[tex]\[ \frac{p}{q} + \frac{r}{q} = \frac{p + r}{q} \][/tex]

### Problem 2: \(\frac{p}{q} - \frac{r}{q}\)

1. Given: Again, we have the fractions \(\frac{p}{q}\) and \(\frac{r}{q}\) where \(q \neq 0\).

2. Common Denominator: Notice that both fractions have the same denominator \(q\).

3. Subtraction: When subtracting fractions with the same denominator, we subtract the numerators while keeping the denominator the same.
[tex]\[ \frac{p}{q} - \frac{r}{q} = \frac{p - r}{q} \][/tex]

Thus,
[tex]\[ \frac{p}{q} - \frac{r}{q} = \frac{p - r}{q} \][/tex]

### Final Results

Putting it all together, we have:

1. \(\frac{p}{q} + \frac{r}{q} = \frac{p + r}{q}\)
2. \(\frac{p}{q} - \frac{r}{q} = \frac{p - r}{q}\)

These are the simplified forms of the given expressions.

Other Questions