SOLUTION: hi im not sure if these maths problems are alebra, eg,1) 2x1/6= 2) 5/14x4 = 3) 7x8/9= please help with these problems im not sure what the formular is to follow to complete these

1) 2× =

We put a 1 under the 2 as a denominator, so that both will be
fractions:

  ×

We observe that the 2 in the numerator and 
the 6 in the denominator can both be divided 
evenly by 2.  

When we divide the 2 by 2 we get 1,
so we cancel the 2 and write a small 1 above it.  

When we divide the 6 by 2 we get 3,
so we cancel the 6 and write a small 3 below it.
Like this:
 
   1
  ×
        3

Next,
we multiply the numbers on top, the little 1 times the 1, getting 1.
we multiply the numbers on the bottom, the 1 times the little 3, getting 3.
and we write the 1 that we got over the 3 and that's the answer:

  

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

2) ×4 =

We put a 1 under the 4 as a denominator, so that both will be
fractions:

  ×

We observe that the 4 in the numerator and 
the 14 in the denominator can both be divided 
evenly by 2.  

When we divide the 4 by 2 we get 2,
so we cancel the 4 and write a small 2 above it.  

When we divide the 14 by 2 we get 7,
so we cancel the 14 and write a small 7 below it.
Like this:
 
         2
  ×
    7

Next,
we multiply the numbers on top, the 5 times the little 2, getting 10.
we multiply the numbers on the bottom, the little 7 times the 1, getting 7.
and we write the 10 that we got over the 7:

  

However, that is an improper fraction because the numerator 10 is larger
than the denominator 7, so we divide the denominator into the numerator,
like this:

   1
7)10
   7
   3

Then we write the quotient 1 as the whole part and after it we write
the fraction  or  and the answer is this
mixed number:



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

3) 7× =

We put a 1 under the 2 as a denominator, so that both will be
fractions:

  ×

This one is different from the other two in that there is no
whole number that either of the numerators can be divided by that
either of the denominators can be divided by to get a smaller number.
So we just multiply the numerators 7×8 and get 56, and multiply the
denominators and get 1×9 or 9, and we put 56 over 9 like this:   

  

However, that is an improper fraction because the numerator 56 is larger
than the denominator 9, so we divide the denominator into the numerator,
like this:

   6
9)56
  54
   2

Then we write the quotient 6 as the whole part and after it we write
the fraction  or  and the answer is this
mixed number:



Edwin