Question 558302
<pre>
{{{3/x}}} + {{{5/y}}}  

3 is the "upper left"
x is the "lower left"
5 is the "upper right"
y is the "lower right"

Rule:

"(Upper left times lower right) plus (upper right times lower left) all over
(bottom times bottom)"

Results:

{{{(3y + 5x)/(xy)}}}

---------------------------------------------------

Same for subtracting, except use "minus"

{{{y/20}}} - {{{10/x}}}  

y is the "upper left"
20 is the "lower left"
10 is the "upper right"
x is the "lower right"

Rule:

"(Upper left times lower right) minus (upper right times lower left) all over
(bottom times bottom)"

Results:

{{{(yx - 10*20)/(20x)}}}

Write 10*20 as 200:

{{{(yx - 200)/(20x)}}}

==========================

Or this second way

{{{3/x}}} + {{{5/y}}}
 
Get the least common denominator of xy

1. Divide the first denominator x into the least common
denominator xy and get y

2. Multiply the first fraction by unit fraction {{{y/y}}}

{{{3/x}}}·{{{y/y}}} + {{{5/y}}}

3. Divide the second denominator y into the least common 
denominator xy and get x

4. Multiply second fraction by unit fraction {{{x/x}}}

{{{3/x}}}·{{{y/y}}} + {{{5/y}}}·{{{x/x}}}

5. Multiply tops and bottoms in both terms

{{{(3y)/(xy)}}} + {{{(5x)/(yx)}}}

6. Since the denominators are the same, (xy and yx are the same),
   add the numerators and put them over the common denominator 

{{{(3y + 5x)/(xy)}}}

Same for subtracting, except use - for +

{{{y/20}}} - {{{10/x}}}
 
Get the least common denominator of 20x

1. Divide the first denominator 20 into the least common
denominator 20x and get x

2. Multiply the first fraction by unit fraction {{{x/x}}}

{{{y/20}}}·{{{x/x}}} - {{{10/x}}}

3. Divide the second denominator x into the least common 
denominator 20x and get 20

4. Multiply second fraction by unit fraction {{{20/20}}}

{{{y/20}}}·{{{x/x}}} - {{{10/x}}}·{{{20/20}}}

5. Multiply tops and bottoms in both terms

{{{(yx)/(20x)}}} - {{{(200)/(20x)}}}

6. Since the denominators are the same,
   subtract the numerators and put them over the common denominator 

{{{(yx - 200)/(20x)}}}

--------------------------------------

Here's a third way to do the first one:

{{{3/x}}} + {{{5/y}}}

1. Get the least common denominator, xy

2. Multiply both fractions by the unit fraction {{{(xy)/(xy)}}}

{{{3/x}}}·{{{(xy)/(xy)}}} + {{{5/y}}}·{{{(xy)/(xy)}}}

3. Cancel left denominators into right numerators in each:

{{{3/cross(x)}}}·{{{(cross(x)y)/(xy)}}} + {{{5/cross(y)}}}·{{{(x*cross(y))/(xy)}}}

{{{(3y)/(xy)}}} + {{{(5x)/(xy)}}}

4. Since the denominators are the same, add the numerators and put them 
   over the common denominator 

{{{(3y + 5x)/(xy)}}}
 
=================================================

I'll do the third one by the second method:

({{{x/20}}} + {{{y/50}}}) - {{{30/y}}}

Remove the parentheses first

{{{x/20}}} + {{{y/50}}} - {{{30/y}}}

Get the least common denominator of 100y

1. Divide the first denominator 20 into the least common
denominator 100y and get 5y

2. Multiply the first fraction by unit fraction {{{(5y)/(5y)}}}

{{{x/20}}}·{{{(5y)/(5y)}}} + {{{y/50}}} - {{{30/y}}}

3. Divide the second denominator 50 into the least common 
denominator 100y and get 2y

4. Multiply second fraction by unit fraction {{{(2y)/(2y)}}}

{{{x/20}}}·{{{(5y)/(5y)}}} + {{{y/50}}}·{{{(2y)/(2y)}}} - {{{30/y}}}

5. Divide the third denominator y into the least common 
denominator 100y and get 100

6. Multiply third fraction by unit fraction {{{100/100}}}

{{{x/20}}}·{{{(5y)/(5y)}}} + {{{y/50}}}·{{{(2y)/(2y)}}} - {{{30/y}}}·{{{100/100}}}

7. Multiply tops and bottoms in all three terms

{{{(5xy)/(100y)}}} + {{{(2y^2)/(100y)}}} - {{{3000/(100y)}}}

8. Since the denominators are the same, combine the numerators and put 
   them over the common denominator 

  {{{(5xy+2y^2-3000)/(100y)}}}


Edwin</pre>