Question

add. write your answer as a fraction or as a whole or mixed number.
$9\frac{3}{11} + 4\frac{6}{11} = \square$

Explanation:

Step1: Add the whole numbers and the fractions separately

For the whole numbers: \(9 + 4 = 13\)
For the fractions: \(\frac{3}{11} + \frac{6}{11} = \frac{3 + 6}{11} = \frac{9}{11}\)

Step2: Combine the results

Now, combine the sum of the whole numbers and the sum of the fractions: \(13 + \frac{9}{11} = 13\frac{9}{11}\) But wait, let's check if the fraction can be simplified or if we can combine further. Wait, actually, let's re - examine the addition of mixed numbers. Another way to add mixed numbers \(a\frac{b}{c}+d\frac{e}{c}\) (where the denominators are the same) is \((a + d)+\frac{b + e}{c}\). Here, \(a = 9\), \(d = 4\), \(b = 3\), \(e = 6\), \(c = 11\). So \((9 + 4)+\frac{3+6}{11}=13+\frac{9}{11}\). But also, we can convert the mixed numbers to improper fractions:
\(9\frac{3}{11}=\frac{9\times11 + 3}{11}=\frac{99+3}{11}=\frac{102}{11}\)
\(4\frac{6}{11}=\frac{4\times11+6}{11}=\frac{44 + 6}{11}=\frac{50}{11}\)
Then add the improper fractions: \(\frac{102}{11}+\frac{50}{11}=\frac{102 + 50}{11}=\frac{152}{11}\)
Now, convert \(\frac{152}{11}\) back to a mixed number: \(152\div11 = 13\) with a remainder of \(9\) (since \(11\times13=143\) and \(152-143 = 9\)), so \(\frac{152}{11}=13\frac{9}{11}\). Wait, but also, let's check the first method again. Wait, actually, if we add the fractions first: \(\frac{3}{11}+\frac{6}{11}=\frac{9}{11}\), and the whole numbers \(9 + 4=13\), so \(13+\frac{9}{11}=13\frac{9}{11}\). But wait, is there a miscalculation? Wait, no, but let's check the sum of the improper fractions: \(102+50 = 152\), and \(152\div11 = 13\) with remainder \(9\), so that's correct. But wait, another way: maybe I made a mistake in the first approach. Wait, no, the two methods are consistent. But wait, let's check the problem again. Wait, \(9\frac{3}{11}+4\frac{6}{11}\). Wait, actually, \(\frac{3 + 6}{11}=\frac{9}{11}\), and \(9 + 4=13\), so \(13\frac{9}{11}\). But also, \(\frac{152}{11}=13\frac{9}{11}\). But wait, let's check if we can simplify \(13\frac{9}{11}\) or if there's an error. Wait, no, that's correct. But wait, wait a second, maybe I made a mistake in the improper fraction conversion. Let's re - do the improper fraction conversion:
For \(9\frac{3}{11}\): \(9\times11=99\), \(99 + 3=102\), so \(\frac{102}{11}\) is correct.
For \(4\frac{6}{11}\): \(4\times11 = 44\), \(44+6 = 50\), so \(\frac{50}{11}\) is correct.
Adding them: \(\frac{102+50}{11}=\frac{152}{11}\). Now, \(11\times13 = 143\), \(152-143 = 9\), so \(\frac{152}{11}=13\frac{9}{11}\). So the answer is \(13\frac{9}{11}\). Wait, but let's check again. Wait, \(9 + 4 = 13\), \(\frac{3}{11}+\frac{6}{11}=\frac{9}{11}\), so \(13+\frac{9}{11}=13\frac{9}{11}\). Yes, that's correct.

Answer:

\(13\frac{9}{11}\)