Question

Solve the Equation: 5/2y+6 + 1/y-2 = 3/y+3

Answer 1

(5/(2y)) + 6 + (1/(y-2) = 3/(y+3) you mean?

Answer 2

No, its a rational expression

Answer 3

\[\frac{5}{2y+6} + \frac{1}{y-2} =\frac{3}{y+3}\]
perhaps?

Answer 4

yes!

Answer 5

ahh k than think carefully, you should make all the denominators equal :)

Answer 6

amistre how you write demoniators and numerators like that o.O? i can't do that -.-'

Answer 7

persistence

Answer 8

:D

Answer 9

so how do I get the answer?

Answer 10

\[\frac{5}{2y+6} + \frac{1}{y-2} -\frac{3}{y+3}=0\]
http://www.wolframalpha.com/input/?i=5%2F%282y%2B6%29+%2B+1%2F%28y-2%29+-3%2F%28y%2B3%29%3D0

Answer 11

there are a lot of ways to get the answer :)

Answer 12

i wish i culd use calculator in algebra exams :P

Answer 13

the main thing to do prolly is just drudge thru the multiplication and simplification ....

Answer 14

clear the fractions one by one by multiplying by each denominator thruout in succession

Answer 15

Im just rying to figure out how to work it out.. i know the answer.. im trying to help a friend.

Answer 16

for example:
\[\frac{2}{A}+\frac{3}{B}-\frac{4}{C}=0\]
\[\frac{2(A)}{A}+\frac{3(A)}{B}-\frac{4(A)}{C}=(A)0\]
\[2(B)+\frac{3A(B)}{B}-\frac{4A(B)}{C}=(B)0\]
\[2B(C)+3A(C)-\frac{4AB(C)}{C}=(C)0\]
\[2BC+3AC-4AB=0\]

Answer 17

then solve for the "y" that is sticking in all the places

Answer 18

blehhh ok i think that sorta helped