Question

do you understand?

1. what is another way to decompose $3\frac{1}{8}$?
2. look at the area model above. what fraction with a greater numerator than denominator is equivalent to $3\frac{1}{8}$? explain.

do you know how?
for 3-4, decompose each fraction or mixed number in two different ways. use a tool if needed.

1. $\frac{3}{5}=\frac{\square}{\square}+\frac{\square}{\square}$ $\frac{3}{5}=\frac{\square}{\square}+\frac{\square}{\square}+\frac{\square}{\square}$
2. $1\frac{3}{4}=1+\frac{\square}{\square}$ $1\frac{3}{4}=\frac{\square}{\square}+\frac{\square}{\square}$

independent practice
leveled practice for 5-10, decompose each fraction or mixed number in two different ways. use a tool if needed.

1. $\frac{4}{6}=$ $\frac{4}{6}=$
2. $\frac{7}{8}=$ $\frac{7}{8}=$
3. $1\frac{3}{5}=$ $1\frac{3}{5}=$
4. $2\frac{1}{2}=$ $2\frac{1}{2}=$
5. $\frac{9}{12}=$ $\frac{9}{12}=$
6. $1\frac{1}{3}=$ $1\frac{1}{3}=$

Answer:

Let's solve these problems one by one. We'll start with problem 3, then 4, and then the independent practice problems (5 - 10).

Problem 3

We need to decompose \(\frac{3}{5}\) in two different ways: as a sum of two fractions and as a sum of three fractions.

First Decomposition (Two Fractions)

We can split the numerator 3 into two parts. Let's take 1 and 2. So:
\[
\frac{3}{5} = \frac{1}{5} + \frac{2}{5}
\]

Second Decomposition (Three Fractions)

We can split the numerator 3 into three parts. Let's take 1, 1, and 1. So:
\[
\frac{3}{5} = \frac{1}{5} + \frac{1}{5} + \frac{1}{5}
\]

Problem 4

We need to decompose \(1\frac{3}{4}\) in two different ways: as \(1 + \text{[fraction]}\) and as a sum of two fractions.

First Decomposition (\(1 + \text{[fraction]}\))

A mixed number \(1\frac{3}{4}\) can be written as the whole number part plus the fractional part. So:
\[
1\frac{3}{4} = 1 + \frac{3}{4}
\]

Second Decomposition (Sum of Two Fractions)

First, convert the mixed number to an improper fraction: \(1\frac{3}{4} = \frac{4\times1 + 3}{4} = \frac{7}{4}\). Now, we can split \(\frac{7}{4}\) into two fractions. Let's take \(\frac{4}{4}\) (which is 1) and \(\frac{3}{4}\), but wait, that's the same as the first decomposition. Wait, no, we can split it differently. Let's take \(\frac{2}{4}\) and \(\frac{5}{4}\)? No, that might not be the best. Wait, actually, \(1\frac{3}{4} = \frac{4}{4} + \frac{3}{4}\) but \(\frac{4}{4}\) is 1. Alternatively, we can split the fractional part. Wait, \(1\frac{3}{4} = \frac{1}{4} + \frac{6}{4}\)? No, that's not helpful. Wait, better: \(1\frac{3}{4} = \frac{4}{4} + \frac{3}{4}\) but \(\frac{4}{4}\) is 1. Wait, maybe \(\frac{5}{4} + \frac{2}{4}\)? Wait, \(\frac{5}{4} + \frac{2}{4} = \frac{7}{4} = 1\frac{3}{4}\). But maybe simpler: \(1\frac{3}{4} = \frac{1}{4} + 1\frac{2}{4}\)? No, that's not right. Wait, the problem says "decompose each fraction or mixed number in two different ways". So for the second way, we can write \(1\frac{3}{4}\) as \(\frac{4}{4} + \frac{3}{4}\) (but \(\frac{4}{4}\) is 1) or as \(\frac{1}{4} + \frac{6}{4}\) but that's not helpful. Wait, actually, \(1\frac{3}{4} = \frac{7}{4}\), so we can split 7 into two parts. Let's take 4 and 3: \(\frac{4}{4} + \frac{3}{4}\) (which is \(1 + \frac{3}{4}\)) or 5 and 2: \(\frac{5}{4} + \frac{2}{4}\). So:
\[
1\frac{3}{4} = \frac{5}{4} + \frac{2}{4}
\]
But \(\frac{5}{4}\) is \(1\frac{1}{4}\) and \(\frac{2}{4}\) is \(\frac{1}{2}\). Alternatively, \(\frac{6}{4} + \frac{1}{4}\) which is \(1\frac{2}{4} + \frac{1}{4} = 1\frac{3}{4}\). So:
\[
1\frac{3}{4} = \frac{6}{4} + \frac{1}{4}
\]

Independent Practice

Problem 5: \(\frac{4}{6}\)

We need to decompose \(\frac{4}{6}\) in two different ways.

First Decomposition

Split the numerator 4 into 1 and 3:
\[
\frac{4}{6} = \frac{1}{6} + \frac{3}{6}
\]

Second Decomposition

Split the numerator 4 into 2 and 2:
\[
\frac{4}{6} = \frac{2}{6} + \frac{2}{6}
\]

Problem 6: \(\frac{7}{8}\)

We need to decompose \(\frac{7}{8}\) in two different ways.

First Decomposition

Split the numerator 7 into 1 and 6:
\[
\frac{7}{8} = \frac{1}{8} + \frac{6}{8}
\]

Second Decomposition

Split the numerator 7 into 3 and 4:
\[
\frac{7}{8} = \frac{3}{8} + \frac{4}{8}
\]

Problem 7: \(1\frac{3}{5}\)

We need to decompose \(1\frac{3}{5}\) in two different ways.

First Decomposition (Whole + Fraction)

\[
1\frac{3}{5} = 1 + \frac{3}{5}
\]

Second Decomposition (Sum of Two Fractions)

Convert to improper fraction: \(1\frac{3}{5} = \frac{5\times1 + 3}{5} = \frac{8}{5}\). Split \(\frac{8}{5}\) into two fractions. Let's take \(\frac{5}{5}\) (which is 1) and \(\frac{3}{5}\), but that's the same as the first. Alternatively, \(\frac{4}{5} + \frac{4}{5}\)? No, \(\frac{4}{5} + \frac{4}{5} = \frac{8}{5}\). Wait, \(\frac{4}{5} + \frac{4}{5} = \frac{8}{5}\), so:
\[
1\frac{3}{5} = \frac{4}{5} + \frac{4}{5}
\]
But \(\frac{4}{5} + \frac{4}{5} = \frac{8}{5} = 1\frac{3}{5}\). Alternatively, \(\frac{2}{5} + \frac{6}{5}\):
\[
1\frac{3}{5} = \frac{2}{5} + \frac{6}{5}
\]

Problem 8: \(2\frac{1}{2}\)

We need to decompose \(2\frac{1}{2}\) in two different ways.

First Decomposition (Whole + Fraction)

\[
2\frac{1}{2} = 2 + \frac{1}{2}
\]

Second Decomposition (Sum of Two Fractions)

Convert to improper fraction: \(2\frac{1}{2} = \frac{2\times2 + 1}{2} = \frac{5}{2}\). Split \(\frac{5}{2}\) into two fractions. Let's take \(\frac{2}{2}\) (which is 1) and \(\frac{3}{2}\), but that's \(1 + \frac{3}{2} = 2\frac{1}{2}\). Alternatively, \(\frac{1}{2} + \frac{4}{2}\):
\[
2\frac{1}{2} = \frac{1}{2} + \frac{4}{2}
\]
But \(\frac{4}{2}\) is 2, so that's \(2 + \frac{1}{2}\) again. Wait, better: \(\frac{3}{2} + \frac{2}{2}\):
\[
2\frac{1}{2} = \frac{3}{2} + \frac{2}{2}
\]
\(\frac{3}{2}\) is \(1\frac{1}{2}\) and \(\frac{2}{2}\) is 1, so \(1\frac{1}{2} + 1 = 2\frac{1}{2}\).

Problem 9: \(\frac{9}{12}\)

We need to decompose \(\frac{9}{12}\) in two different ways.

First Decomposition

Split the numerator 9 into 3 and 6:
\[
\frac{9}{12} = \frac{3}{12} + \frac{6}{12}
\]
Simplify: \(\frac{3}{12} = \frac{1}{4}\) and \(\frac{6}{12} = \frac{1}{2}\), but we can keep it as \(\frac{3}{12} + \frac{6}{12}\) or simplify.

Second Decomposition

Split the numerator 9 into 4 and 5:
\[
\frac{9}{12} = \frac{4}{12} + \frac{5}{12}
\]
Simplify: \(\frac{4}{12} = \frac{1}{3}\) and \(\frac{5}{12}\) remains, but we can keep it as \(\frac{4}{12} + \frac{5}{12}\).

Problem 10: \(1\frac{1}{3}\)

We need to decompose \(1\frac{1}{3}\) in two different ways.

First Decomposition (Whole + Fraction)

\[
1\frac{1}{3} = 1 + \frac{1}{3}
\]

Second Decomposition (Sum of Two Fractions)

Convert to improper fraction: \(1\frac{1}{3} = \frac{3\times1 + 1}{3} = \frac{4}{3}\). Split \(\frac{4}{3}\) into two fractions. Let's take \(\frac{3}{3}\) (which is 1) and \(\frac{1}{3}\), but that's the same as the first. Alternatively, \(\frac{2}{3} + \frac{2}{3}\):
\[
1\frac{1}{3} = \frac{2}{3} + \frac{2}{3}
\]
But \(\frac{2}{3} + \frac{2}{3} = \frac{4}{3} = 1\frac{1}{3}\). Alternatively, \(\frac{1}{3} + \frac{3}{3}\):
\[
1\frac{1}{3} = \frac{1}{3} + \frac{3}{3}
\]
\(\frac{3}{3}\) is 1, so that's \(1 + \frac{1}{3}\) again. Wait, better: \(\frac{4}{3} = \frac{5}{3} - \frac{1}{3}\)? No, we need to decompose into a sum. Wait, \(1\frac{1}{3} = \frac{4}{3} = \frac{1}{3} + \frac{3}{3}\) (which is \(1 + \frac{1}{3}\)) or \(\frac{2}{3} + \frac{2}{3}\). So:
\[
1\frac{1}{3} = \frac{2}{3} + \frac{2}{3}
\]

Final Answers

Problem 3

\[
\frac{3}{5} = \frac{1}{5} + \frac{2}{5} \quad \text{and} \quad \frac{3}{5} = \frac{1}{5} + \frac{1}{5} + \frac{1}{5}
\]

Problem 4

\[
1\frac{3}{4} = 1 + \frac{3}{4} \quad \text{and} \quad 1\frac{3}{4} = \frac{4}{4} + \frac{3}{4} \quad (\text{or} \quad \frac{5}{4} + \frac{2}{4})
\]

Problem 5

\[
\frac{4}{6} = \frac{1}{6} + \frac{3}{6} \quad \text{and} \quad \frac{4}{6} = \frac{2}{6} + \frac{2}{6}
\]

Problem 6

\[
\frac{7}{8} = \frac{1}{8} + \frac{6}{8} \quad \text{and} \quad \frac{7}{8} = \frac{3}{8} + \frac{4}{8}
\]

Problem 7

\[
1\frac{3}{5} = 1 + \frac{3}{5} \quad \text{and} \quad 1\frac{3}{5} = \frac{4}{5} + \frac{4}{5} \quad (\text{or} \quad \frac{2}{5} + \frac{6}{5})
\]

Problem 8

\[
2\frac{1}{2} = 2 + \frac{1}{2} \quad \text{and} \quad 2\frac{1}{2} = \frac{3}{2} + \frac{2}{2} \quad (\text{or} \quad \frac{1}{2} + \frac{4}{2})
\]

Problem 9

\[
\frac{9}{12} = \frac{3}{12} + \frac{6}{12} \quad \text{and} \quad \frac{9}{12} = \frac{4}{12} + \frac{5}{12}
\]

Problem 10

\[
1\frac{1}{3} = 1 + \frac{1}{3} \quad \text{and} \quad 1\frac{1}{3} = \frac{2}{3} + \frac{2}{3} \quad (\text{or} \quad \frac{1}{3} + \frac{3}{3})
\]