Question

subtract. answer as a mixed number in lowest terms. \\(\

$$\begin{array}{r}4\\frac{1}{4}\\\\ - 1\\frac{3}{8}\\\\ \\hline 2\\frac{?}{\\square}\\end{array}$$

\\) next, enter the numerator.

Explanation:

Step1: Convert mixed numbers to improper fractions

First, convert \(4\frac{1}{4}\) and \(1\frac{3}{8}\) to improper fractions.
For \(4\frac{1}{4}\), we use the formula \(a\frac{b}{c}=\frac{a\times c + b}{c}\). So, \(4\frac{1}{4}=\frac{4\times4 + 1}{4}=\frac{17}{4}\).
For \(1\frac{3}{8}\), we have \(1\frac{3}{8}=\frac{1\times8+3}{8}=\frac{11}{8}\).

Step2: Find a common denominator

The denominators are 4 and 8. The least common denominator (LCD) of 4 and 8 is 8. So, we convert \(\frac{17}{4}\) to a fraction with denominator 8.
To do this, we multiply the numerator and denominator of \(\frac{17}{4}\) by 2: \(\frac{17\times2}{4\times2}=\frac{34}{8}\).

Step3: Subtract the fractions

Now, subtract \(\frac{11}{8}\) from \(\frac{34}{8}\): \(\frac{34}{8}-\frac{11}{8}=\frac{34 - 11}{8}=\frac{23}{8}\).

Step4: Convert back to a mixed number

Convert \(\frac{23}{8}\) back to a mixed number. Divide 23 by 8: \(23\div8 = 2\) with a remainder of 7. So, \(\frac{23}{8}=2\frac{7}{8}\). Wait, but the problem shows the whole number part as 2 already. Wait, maybe I made a mistake in the initial whole number subtraction? Wait, let's re - examine. Wait, maybe the problem has a typo or maybe I misread. Wait, no, let's do the subtraction of mixed numbers directly.

Alternative approach: Subtract the whole numbers and the fractions separately, but we need to borrow if the fraction part of the minuend is less than the fraction part of the subtrahend.

The mixed numbers are \(4\frac{1}{4}\) and \(1\frac{3}{8}\). The fraction parts are \(\frac{1}{4}\) and \(\frac{3}{8}\). Convert \(\frac{1}{4}\) to eighths: \(\frac{1}{4}=\frac{2}{8}\). Now, \(\frac{2}{8}<\frac{3}{8}\), so we need to borrow 1 from the whole number part of \(4\frac{1}{4}\).

Borrowing 1 (which is \(\frac{8}{8}\)) from 4, we get \(3 + 1+\frac{1}{4}=3+\frac{8}{8}+\frac{2}{8}=3\frac{10}{8}\) (wait, \(\frac{1}{4}=\frac{2}{8}\), so \(4\frac{1}{4}=3 + 1+\frac{1}{4}=3+\frac{8}{8}+\frac{2}{8}=3\frac{10}{8}\)? No, \(4\frac{1}{4}=3 + 1\frac{1}{4}=3+\frac{5}{4}=3\frac{5}{4}\)? Wait, no, the correct way to borrow is: \(4\frac{1}{4}=3 + 1+\frac{1}{4}=3+\frac{4}{4}+\frac{1}{4}=3\frac{5}{4}\)? No, I'm getting confused. Let's use the first method. Wait, when we did the improper fraction method, we got \(\frac{23}{8}=2\frac{7}{8}\), but the problem shows the whole number as 2. Wait, maybe the original problem's whole number subtraction is 4 - 1 - 1=2? Wait, 4 - 1=3, but we borrowed 1, so 3 - 1=2. Then the fraction part: after borrowing, we had \(\frac{1}{4}+\frac{4}{4}=\frac{5}{4}=\frac{10}{8}\), then \(\frac{10}{8}-\frac{3}{8}=\frac{7}{8}\). So the mixed number is \(2\frac{7}{8}\). So the numerator is 7.

Wait, let's check the subtraction again:

\(4\frac{1}{4}-1\frac{3}{8}\)

Convert \(4\frac{1}{4}\) to \(3\frac{5}{4}\) (because we need to subtract the fraction \(\frac{3}{8}\) from \(\frac{1}{4}\), but \(\frac{1}{4}<\frac{3}{8}\), so we borrow 1 from 4, making it 3, and add that 1 (which is \(\frac{4}{4}\)) to \(\frac{1}{4}\), getting \(\frac{5}{4}\)). Now, convert \(\frac{5}{4}\) to eighths: \(\frac{5}{4}=\frac{10}{8}\).

Now subtract the whole numbers: 3 - 1 = 2.

Subtract the fractions: \(\frac{10}{8}-\frac{3}{8}=\frac{7}{8}\).

So the result is \(2\frac{7}{8}\), so the numerator is 7.

Answer:

7