Adding and Subtracting Unlike Denominators (Help)

\(\text{the key is to get them over a single denominator}\\ \dfrac 2 5 - \dfrac{3}{3y} = \\ \dfrac{2\cdot 3y - 3\cdot 5}{5\cdot 3y}=\\ \dfrac{6y-15}{15y} = \\ \dfrac{2y-5}{5y} =\\ \text{and you can go one step further if you like}\\ \dfrac 2 5 - \dfrac 1 y\)

Thanks, Rom.....

Here's another proceedure that always works, BUT....it may NOT produce a fraction in "reduced" form....however...if you are good at reducing "final" fractions....it relieves you of the task of finding the "common denominator"

Note....suppose that we have

  a    -    c

__        __

 b          d

Reduce any fractions that we can, first

Ctross multiply  in this order    ⇒   ad    and  bc    

Seperate these with the same sign as we have between the fractions

So we have   ad  -  bc

Put this over the product of the denominators  =  bd

So..we have

ad  -  bc

______

   bd

So...in your problem, we have

2       -       3

__           ___

5               3y

Note that the second fraction can be reduced first... so we have

2     -       1

__         ___

5             y   

Cross-multiply in the specified order

2*y  - 5*1     =      2y - 5

Product of the denominators  

5 * y    =   5y

So  we have

2y - 5

______       [ which is the same thing as Rom's 4th step  !! ]

   5y