fraction problem?

Question

1/3 + 1/y
-------------- (a fraction)
y/3 – 3/y

Answer 1

well dear Schnitzel

The first thing we should do is finding the smallest, or least, common denominator Denominator The least common denominator , or LCD, of two fractions is the smallest number that can be divided by both denominators.Now just multiply denominators
(1/3 + 1/y) / ( y/3 – 3/y )

• Step 1
1/3 + 1/y
common denominator is 3 * y = 3y
((1*y ) + ( 1 * 3)) / 3y =
(y + 3) / 3y

• Step 2
y/3 – 3/y
common denominator is 3 * y = 3y
((y * y) - ( 3 *3)) / 3y =
( y^2 - 9 ) / 3y

• Step 3
((y + 3) / 3y) * (( y^2 - 9 ) / 3y )

{ far * far & near * near }
((y + 3) * 3y ) / (( y^2 - 9 ) * 3y )
Simplify " 3y "
(y + 3) / ( y^2 - 9 )
{ as you know y^2 - 9 = ( y -3)( y+3)
so we have ;
(y + 3) / ( ( y -3)( y+3) )
Simplify " y + 3 "

so we have ;
1/ ( y - 3)

Good Luck Dear...

Answer 2

Simplify?

(1/3 + 1/y)/(y/3 – 3/y)

The LCD of the little fractions is 3y, so multiply the big numerator and the big denominator by 3y to clear the little fractions

[3y(1/3 + 1/y)]/[3y(y/3 – 3/y)]
(y + 3)/(y^2 - 9)

The denominator factors (difference of two squares)

(y + 3)/[(y + 3)(y - 3)]

1/(y - 3)

Answer 3

First of all, combine like fractions by finding a common denominator.

Top half: common denominator is 3y
y/3y + 3/3y = (y+3)/(3y)

Bottom half: common denominator is also 3y
y^2/3y - 9/3y = (y^2-9)/(3y)

Now, to divide fractions, you multiply the top half by the reciprocal of the bottom half:

(y+3)/(3y) * (3y)/(y^2-9)

The two "3y" cancel each other out:

(y+3)/(y^2-9)

The factor "(y^2-9)" factors into "(y+3)" and "(y-3)". The "(y+3)" cancel out leaving:

1/(y-3)

Answer 4

1/(y-3)

Answer 5

1/(y-3)

Answer 6

LCD numerator = LCD denominator = 3y

Mutiply fraction by

3y
---
3y

And you get

1/3 + 1/y ... 3y ........ y + 3 ......... (y + 3) ............... 1
------------ X ---...=...--------- ... = ---------------- ... = ---------
y/3 – 3/y .... 3y ....... y² - 9 ...... (y + 3)(y - 3) ......... y - 3

(NOTE y² - 9 = y² - 3² = (y + 3)(y - 3)

Answer 7

maybe u should get a tutor or a teacher 2 help u bc if you ask ppl 4 the answers you'll never learn. good luck! ;-)

Answer 8

(y+3)/3y *3y/(y+3)(y-3)
=1/(y-3)

Answer 9

(y+3)/3y
----------- = ((y+3)/3y)*(3y/(y^2-9))= 1/(y-3)
(y^2-9)/3y