# How do you evaluate 16\frac { 2} { 7} - 2\frac { 1} { 2} \cdot 1\frac { 1} { 7} + \frac { 9} { 14}?

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Feb 11, 2017

$16 \frac{2}{7} - 2 \frac{1}{2} \cdot 1 \frac{1}{7} + \frac{9}{14} = 14 \frac{1}{14}$

#### Explanation:

Given:

$16 \frac{2}{7} - 2 \frac{1}{2} \cdot 1 \frac{1}{7} + \frac{9}{14}$

First let us convert all of the mixed numbers to improper fractions:

$16 \frac{2}{7} = 16 + \frac{2}{7} = \frac{16 \cdot 7}{7} + \frac{2}{7} = \frac{112}{7} + \frac{2}{7} = \frac{114}{7}$

$2 \frac{1}{2} = 2 + \frac{1}{2} = \frac{2 \cdot 2}{2} + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$

$1 \frac{1}{7} = 1 + \frac{1}{7} = \frac{7}{7} + \frac{1}{7} = \frac{8}{7}$

So our original expression can be rewritten as:

$\frac{114}{7} - \frac{5}{2} \cdot \frac{8}{7} + \frac{9}{14}$

Next note that multiplication has higher precedence than addition or multiplication, so we need to perform the multiplication $\frac{5}{2} \cdot \frac{8}{7}$ first:

$\frac{114}{7} - \textcolor{b l u e}{\frac{5}{2} \cdot \frac{8}{7}} + \frac{9}{14} = \frac{114}{7} - \frac{5 \cdot 8}{2 \cdot 7} + \frac{9}{14} = \frac{114}{7} - \frac{40}{14} + \frac{9}{14}$

In order to add or subtract these fractions, they need to have the same denominators, so we multiply $\frac{114}{7}$ by $\frac{2}{2}$ to give it a denominator $14$ like the other fractions:

$\frac{114}{7} = \frac{114 \cdot 2}{7 \cdot 2} = \frac{228}{14}$

So our express can be rewritten:

$\frac{228}{14} - \frac{40}{14} + \frac{9}{14} = \frac{228 - 40 + 9}{14}$

Then note that addition and subtraction have the same priority, so we need to evaluate them from left to right:

$\frac{\textcolor{b l u e}{228 - 40} + 9}{14} = \frac{188 + 9}{14} = \frac{197}{14}$

To express this as a mixed number, we divide $197$ by $14$ to give a quotient $14$ and remainder $1$, so:

$\frac{197}{14} = 14 + \frac{1}{14} = 14 \frac{1}{14}$

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Feb 12, 2017

$= 14 \frac{1}{14}$

#### Explanation:

It is possible to work with the whole numbers separately when adding or subtracting.

Only for multiplication do we have to use improper fractions.

There are 3 terms, do the multiplication first.

$16 \frac{2}{7} \textcolor{b l u e}{- 2 \frac{1}{2} \times 1 \frac{1}{7}} + \frac{9}{14}$

$= 16 \frac{2}{7} \textcolor{b l u e}{- \frac{5}{2} \times \frac{8}{7}} + \frac{9}{14}$

We would usually cancel in the middle term. However we will need a common denominator in the next step so leave as it is. Simplify by multiplying straight across.

$16 \frac{4}{14} \textcolor{b l u e}{- \frac{40}{14}} + \frac{9}{14} \text{ }$ all have the same denominator

$= 16 \frac{4}{14} \textcolor{b l u e}{- 2 \frac{12}{14}} + \frac{9}{14} \text{ }$ change to a mixed number

$= 14 \frac{4 - 12 + 9}{14}$

$= 14 \frac{1}{14}$

Working with mixed numbers gives us smaller numbers to work with. The numbers in improper fractions are often uncomfortably big.

(Answer in the same form as the numbers were given.)

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Feb 11, 2017

$14 \frac{1}{14}$

#### Explanation:

$16 \frac{2}{7} - 2 \frac{1}{2} \cdot 1 \frac{1}{7} + \frac{9}{14}$

$\therefore = \frac{114}{7} - \left(\frac{5}{2} \times \frac{8}{7}\right) + \frac{9}{14}$

$\therefore = \frac{114}{7} - \left(\frac{40}{14}\right) + \frac{9}{14}$

$\therefore = \frac{228 - 40 + 9}{14}$

$\therefore = \frac{197}{14}$

$\therefore = 14 \frac{1}{14}$

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