#1**+2 **

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

.Rom Nov 5, 2018

#3**+2 **

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

CPhill Nov 5, 2018