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# math help subtracting mixed numbers

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42 1/6-27 3/5.

Nov 4, 2018

#1
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Okay I misread the problem. Alright.

$$42\frac{1}{6} - 27\frac{3}{5}$$

First off, we want to convert these mixed numbers into improper fractions.

We get $$\frac{42\times6 + 1}{6} - \frac{27\times5 + 3}{5}$$, or $$\frac{253}{6} - \frac{138}{5}$$.

We need to put both fractions under a common denominator to solve. Let's put both fractions under $$30$$, the LCM.

We get $$\frac{1265}{30} - \frac{828}{30}$$. Now we can subtract the fractions, giving us $$\frac{437}{30}$$.

Converting this back into a mixed number, we get $$14\frac{17}{30}$$.

Therefore, our answer is $$\boxed{14\frac{17}{30}}$$.

Nov 4, 2018
edited by KnockOut  Nov 4, 2018
edited by KnockOut  Nov 4, 2018
#2
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they are mixed numbers  just a subtraction symbol

Nov 4, 2018
#3
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An easier way to do this is to subtract the mixed numbers without conversion to an improper fraction. For this to happen, the fraction part of the mixed number in the first number ($$\frac{1}{6}$$ in our case) needs to be larger than the $$\frac{3}{5}$$.

Let's compare.

$$\frac{1}{6} = \frac{5}{30}$$

$$\frac{3}{5} = \frac{18}{30}$$

$$\frac{18}{30}>\frac{5}{30}$$.

Since the second fraction is larger, we need to modify the first fraction to make it larger.

So from $$42\frac{1}{6}$$, we are going to take $$1$$ (Which is equal to $$\frac{6}{6}$$) out of the $$42$$, and add it to the fractions part giving us $$41\frac{7}{6}$$

$$\frac{7}{6} = \frac{35}{30}$$.

$$\frac{35}{30}>\frac{18}{30}$$

Now we can subtract easily.

$$41\frac{35}{30} - 27\frac{18}{30} = 14\frac{17}{30}$$

So our answer is yet again $$\boxed{14\frac{17}{30}}$$

Nov 4, 2018