Algbrea 2 Math Could someone please help me

Question

3y/over y(5y-1)  + 2/over 5y-1=

 

Answer in lowest terms and leave in factored form.

Answer 1

Fast method:

Cancel out the y's in the first fraction:

3y           +     2
----              ------- = ?
y(5y-1)          5y-1


Just add, since the denominator's are the same:
3+2
-----
5y-1

5
---
5y-1

 

 

 

More involved approach:

 

 

3y          +     2
----            ------- = ?

y(5y-1)         5y-1

 

 

The LCM of y(5y-1) and (5y-1) is:  y(5y-1)

 

 

Note that is already the denominator for the fraction on the left. 

All you need to do to get a common denominator is multiply the numerator and the denominator of the 2nd fraction by y:

 

 

3y              +    2(y)

----                   --------

y(5y-1)              y (5y-1)

 

 

Since they now have a common denominator:

 

3y + 2y

--------

y(5y-1)

 

 

5y

------

y(5y-1)

 

Cancel the y's.

 

5

----

5y-1

 

 

 

 

Answer 2

3y    +    2

----       ------- = ?

y(5y-1)    5y-1

 

The simplest thing to do is to multiply by the reciprocal of the denominators

 

y*(5*y-1)*3*y  +   y*(5*y-1)*2

------------------       --------------- = ?*y*(5*y-1)

y*(5*y-1)                  (5*y-1)

 

Cancel the factors that are in both the denominator and numerator for each fraction

 

3*y  + y*2 = ?*y*(5*y-1)

 

Simplify

5*y = ?*y*(5*y-1)

 

Without the "?" filled in, it's hard to tell how far to go, but it does say leave in factored form, so you shouldn't multiply through by the y.

 

On the other hand, you could divide both sides by (5*y)

 

5*y       ?*y*(5*y-1)

----   = ----------------

5*y           5*y

 

leaving

 

1 = ?*(1/5)*(5*y-1) = ?*(y - (1/5))

 

Which is pretty simplified.