1. ## subtract

PROBLEM: 6/8 - 2/6 =

10 1/4 + 13 3/6=

2. Originally Posted by breanna2471

PROBLEM: 6/8 - 2/6 =

10 1/4 + 13 3/6=
You need to get a common denominator. The most "efficient" denominator is the Least Common Multiple (LCM) of the two denominators, but if you can't figure out what that might be you can simply use the product of the two denominators.

$\frac{6}{8} - \frac{2}{6}$
Frankly I'd express these in lowest terms before I subtracted. This isn't necessary, but makes the numbers a little nicer. So....
$\frac{6}{8} - \frac{2}{6} = \frac{3}{4} - \frac{1}{3}$

The LCM of 4 and 3 is 12. So we want the denominator in each fraction to become 12. For the $\frac{3}{4}$ that means multiplying the denominator by 3. But what we do in the denominator we must do in the numerator as well:
$\frac{3}{4} = \frac{3}{4} \cdot \frac{3}{3} = \frac{9}{12}$

Similarly for $\frac{1}{3}$ we need to mulitply the denominator by 4:
$\frac{1}{3} = \frac{1}{3} \cdot \frac{4}{4} = \frac{4}{12}$.

Thus
$\frac{6}{8} - \frac{2}{6} = \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12}$

$= \frac{9 - 4}{12} = \frac{5}{12}$
====================================
$10 ~ \frac{1}{4} + 13 ~ \frac{3}{6}$

I'm going to reorder this so that I'm adding whole numbers to whole numbers and adding fractions to fractions. (Or you could just express these as improper fractions and add the fractions.)
$10 ~ \frac{1}{4} + 13 ~ \frac{3}{6} = (10 + 13) + \left ( \frac{1}{4} + \frac{3}{6} \right )$

Again I'm going to put $\frac{3}{6} = \frac{1}{2}$ in lowest terms:
$23 + \left ( \frac{1}{4} + \frac{1}{2} \right )$

Now add the fractions. The LCM of 4 and 2 is 4. Thus we need do nothing to the first fraction. To make the denominator in the second fraction 4 we multiply it by 2. So:
$\frac{1}{2} = \frac{1}{2} \cdot \frac{2}{2} = \frac{2}{4}$

Thus
$10 ~ \frac{1}{4} + 13 ~ \frac{3}{6} = 23 + \left ( \frac{1}{4} + \frac{2}{4} \right )$

$= 23 + \left ( \frac{1 + 2}{4} \right )$

$= 23 + \left ( \frac{3}{4} \right )$

or
$23 ~ \frac{3}{4}$

-Dan

3. ## How to solve fractions ...

Originally Posted by breanna2471

PROBLEM: 6/8 - 2/6 =

10 1/4 + 13 3/6=

Problem
$
6/8 - 2/6
$

If the above problem was just 6 - 2, you'd answer 4 (won't need our help )

If it was 6/1 - 2/1, you'd still answer 4/1 or 4 (won't need our help again )

But why ?

Because the denominator was 1 for both the numbers. This made the problem very easy!

$
\frac{6}{8} - \frac{2}{6}
$

The denominators are not the same... We can't simply subtract like we did before...

So... lets make the denominators the same by multiplying and dividing by the number that's missing

$
\frac{6}{8} - \frac{2}{6}
= \frac{6}{8}*\frac{6}{6} - \frac{2}{6}*\frac{8}{8}
$

(Note that in the above, I multiplied and divided by 6 for one number, and by 8 for the other number. By doing so, I did not change the value of the expression)

Now, we have
$
\frac{6}{8}*\frac{6}{6} - \frac{2}{6}*\frac{8}{8}
= \frac{6*6}{8*6} - \frac{2*8}{6*8}
= \frac{36}{48} - \frac{16}{48}
$

Since the denominator is 48 in both the cases, we can proceed to simplify it as follows

$
= \frac{36-16}{48}
= \frac{20}{48}
= \frac{5}{12}
$

Hope this helps

Mahurshi Akilla

4. Originally Posted by breanna2471
$\frac34-\frac13=\frac{12}{12}\cdot\left(\frac34-\frac13\right)=\frac1{12}\cdot(9-4)=\frac5{12}$