Question

subtract the mixed numbers. simplify your answer if needed.
1 ( 3\frac{1}{3} - 1\frac{2}{3} )
2 ( 6\frac{1}{4} - 3\frac{3}{4} )
3 ( 5\frac{3}{8} - 1\frac{7}{8} )
4 ( 2\frac{4}{7} - 1\frac{6}{7} )
5 ( 8\frac{4}{10} - 2\frac{7}{10} )
6 ( 3\frac{3}{6} - 1\frac{5}{6} )
7 ( 8\frac{2}{14} - 5\frac{9}{14} )
8 ( 5\frac{3}{16} - 1\frac{7}{16} )
9 ( 9\frac{2}{9} - 5\frac{5}{9} )
10 ( 4\frac{2}{11} - 2\frac{6}{11} )
11 ( 7\frac{1}{7} - 5\frac{5}{7} )
12 ( 2\frac{5}{12} - 1\frac{8}{12} )

Explanation:

Step1: Subtract the fractional parts

For \(3\frac{1}{3}-1\frac{2}{3}\), the fractional parts are \(\frac{1}{3}\) and \(\frac{2}{3}\). Since \(\frac{1}{3}<\frac{2}{3}\), we need to borrow 1 from the whole number part of \(3\frac{1}{3}\). So \(3\frac{1}{3}=2 + 1+\frac{1}{3}=2+\frac{3}{3}+\frac{1}{3}=2\frac{4}{3}\).

Step2: Subtract the mixed numbers

Now subtract the whole numbers and the fractions separately: \(2\frac{4}{3}-1\frac{2}{3}=(2 - 1)+(\frac{4}{3}-\frac{2}{3}) = 1+\frac{2}{3}=1\frac{2}{3}\)? Wait, no, wait. Wait, actually, when subtracting mixed numbers with like denominators, if the numerator of the subtrahend's fraction is greater than the minuend's fraction numerator, we borrow 1 from the whole number as a fraction. But let's do it correctly. \(3\frac{1}{3}-1\frac{2}{3}\), rewrite \(3\frac{1}{3}\) as \(2\frac{4}{3}\) (because \(3 = 2 + 1\) and \(1=\frac{3}{3}\), so \(3\frac{1}{3}=2+\frac{3}{3}+\frac{1}{3}=2\frac{4}{3}\)). Then subtract: \(2\frac{4}{3}-1\frac{2}{3}=(2 - 1)+(\frac{4}{3}-\frac{2}{3})=1+\frac{2}{3}=1\frac{2}{3}\)? Wait, no, that's not right. Wait, \(3\frac{1}{3}-1\frac{2}{3}\), the whole number part: \(3 - 1 = 2\), fraction part: \(\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}\), so we have \(2-\frac{1}{3}=1\frac{2}{3}\). Yes, that's correct.

Wait, maybe a better way: For mixed numbers with the same denominator, \(a\frac{b}{c}-d\frac{e}{c}=(a - d)+\frac{b - e}{c}\), if \(b\geq e\), otherwise, \( (a - 1 - d)+\frac{b + c - e}{c}\).

So for problem 1: \(3\frac{1}{3}-1\frac{2}{3}\), \(b = 1\), \(e = 2\), so \(b

Problem 2: \(6\frac{1}{4}-3\frac{3}{4}\), \(b = 1\), \(e = 3\), \(b

Problem 3: \(5\frac{3}{8}-1\frac{7}{8}\), \(b = 3\), \(e = 7\), \(b

Problem 4: \(2\frac{4}{7}-1\frac{6}{7}\), \(b = 4\), \(e = 6\), \(b

Problem 5: \(8\frac{4}{10}-2\frac{7}{10}\), \(b = 4\), \(e = 7\), \(b

Problem 6: \(3\frac{3}{6}-1\frac{5}{6}\), \(b = 3\), \(e = 5\), \(b

Problem 7: \(8\frac{2}{14}-5\frac{9}{14}\), \(b = 2\), \(e = 9\), \(b

Problem 8: \(5\frac{3}{16}-1\frac{7}{16}\), \(b = 3\), \(e = 7\), \(b

Problem 9: \(9\frac{2}{9}-5\frac{5}{9}\), \(b = 2\), \(e = 5\), \(b

Problem 10: \(4\frac{2}{11}-2\frac{6}{11}\), \(b = 2\), \(e = 6\), \(b

Problem 11: \(7\frac{1}{7}-5\frac{5}{7}\), \(b = 1\), \(e = 5\), \(b

Problem 12: \(2\frac{5}{12}-1\frac{8}{12}\), \(b = 5\), \(e = 8\), \(b

Answer:

s (for each problem):

1. \(1\frac{2}{3}\)
2. \(2\frac{1}{2}\)
3. \(3\frac{1}{2}\)
4. \(\frac{5}{7}\)
5. \(5\frac{7}{10}\)
6. \(1\frac{2}{3}\)
7. \(2\frac{1}{2}\)
8. \(3\frac{3}{4}\)
9. \(3\frac{2}{3}\)
10. \(1\frac{7}{11}\)
11. \(1\frac{3}{7}\)
12. \(\frac{3}{4}\)