Question

operations with mixed numbers
subtract: $2\frac{1}{4}-1\frac{2}{3}$
options: $1\frac{7}{12}$, $\frac{1}{3}$, $\frac{5}{12}$, $\frac{7}{12}$

Explanation:

Step1: Convert mixed numbers to improper fractions

To subtract mixed numbers \(2\frac{1}{4}-1\frac{2}{3}\), first convert them to improper fractions.

For \(2\frac{1}{4}\), the formula to convert a mixed number \(a\frac{b}{c}\) to an improper fraction is \(\frac{a\times c + b}{c}\). So, \(2\frac{1}{4}=\frac{2\times4 + 1}{4}=\frac{8 + 1}{4}=\frac{9}{4}\).

For \(1\frac{2}{3}\), using the same formula, \(1\frac{2}{3}=\frac{1\times3+2}{3}=\frac{3 + 2}{3}=\frac{5}{3}\).

Step2: Find a common denominator

The denominators are 4 and 3. The least common denominator (LCD) of 4 and 3 is \(4\times3 = 12\).

Rewrite \(\frac{9}{4}\) and \(\frac{5}{3}\) with the denominator 12:

\(\frac{9}{4}=\frac{9\times3}{4\times3}=\frac{27}{12}\)

\(\frac{5}{3}=\frac{5\times4}{3\times4}=\frac{20}{12}\)

Step3: Subtract the fractions

Now subtract the two fractions: \(\frac{27}{12}-\frac{20}{12}=\frac{27 - 20}{12}=\frac{7}{12}\)? Wait, no, wait. Wait, we made a mistake here. Wait, the original mixed numbers: \(2\frac{1}{4}\) is \(2 + \frac{1}{4}\), and \(1\frac{2}{3}\) is \(1+\frac{2}{3}\). Maybe a better way is to handle the whole numbers and fractions separately, but we have to borrow if the fraction part of the minuend is less than the fraction part of the subtrahend.

Let's try again. \(2\frac{1}{4}=1 + 1+\frac{1}{4}=1+\frac{4}{4}+\frac{1}{4}=1+\frac{5}{4}\)? No, better to convert to improper fractions correctly. Wait, \(2\frac{1}{4}=\frac{9}{4}\), \(1\frac{2}{3}=\frac{5}{3}\). Now, find LCD of 4 and 3, which is 12. So \(\frac{9}{4}=\frac{27}{12}\), \(\frac{5}{3}=\frac{20}{12}\). Then \(\frac{27}{12}-\frac{20}{12}=\frac{7}{12}\)? But wait, that can't be, because \(2\frac{1}{4}\) is about 2.25, and \(1\frac{2}{3}\) is about 1.666..., so the difference should be about 0.583..., and \(\frac{7}{12}\approx0.583\), which is correct. Wait, but let's check again.

Wait, \(2\frac{1}{4}-1\frac{2}{3}\). Let's do it by separating whole numbers and fractions:

\(2\frac{1}{4}=2+\frac{1}{4}\)

\(1\frac{2}{3}=1+\frac{2}{3}\)

So, \(2 - 1=1\), and \(\frac{1}{4}-\frac{2}{3}\). But \(\frac{1}{4}<\frac{2}{3}\), so we need to borrow 1 from the whole number part of the minuend. So, \(2\frac{1}{4}=1 + 1+\frac{1}{4}=1+\frac{4}{4}+\frac{1}{4}=1+\frac{5}{4}\). Now, subtract:

\(1+\frac{5}{4}-(1+\frac{2}{3})=1 - 1+\frac{5}{4}-\frac{2}{3}=0+\frac{15}{12}-\frac{8}{12}=\frac{7}{12}\)? Wait, no, \(\frac{5}{4}-\frac{2}{3}=\frac{15 - 8}{12}=\frac{7}{12}\). Wait, but that gives \(\frac{7}{12}\), but let's check the options. Wait, the options include \(\frac{5}{12}\), \(\frac{7}{12}\), etc. Wait, maybe I made a mistake in the borrowing.

Wait, let's recalculate:

\(2\frac{1}{4}=\frac{9}{4}\), \(1\frac{2}{3}=\frac{5}{3}\)

Find a common denominator, which is 12.

\(\frac{9}{4}=\frac{9\times3}{4\times3}=\frac{27}{12}\)

\(\frac{5}{3}=\frac{5\times4}{3\times4}=\frac{20}{12}\)

Subtract: \(\frac{27}{12}-\frac{20}{12}=\frac{7}{12}\). Wait, but the options have \(\frac{5}{12}\), \(\frac{7}{12}\), etc. Wait, maybe I messed up the mixed number conversion. Wait, \(2\frac{1}{4}\) is 2.25, \(1\frac{2}{3}\) is approximately 1.666..., so 2.25 - 1.666... = 0.583..., and \(\frac{7}{12}\approx0.583\), which is correct. So the answer should be \(\frac{7}{12}\)? Wait, but let's check the options again. The options are:

• \(1\frac{7}{12}\)

• \(\frac{1}{3}\)

• \(\frac{5}{12}\)

• \(\frac{7}{12}\)

Wait, so \(\frac{7}{12}\) is one of the options. So that's the correct answer.

Answer:

\(\frac{7}{12}\)