Question

How do I solve this?

1 + (1 / y) = (6 / y^2)

Answer 1

Do you mean this: \[1 +\frac{ 1 }{ y } = \frac{ 6 }{ y^{2} }\]

Answer 2

yup!

Answer 3

First take LCM. of Left hand side. like. \[1 + \frac{ 1 }{ 3 } = \frac{ 4 }{ 3 }\]

Answer 4

multiply with y^2 what do you get?

Answer 5

hmm...

(y^3 / y^3) + (y^2 / y^3) = (6y / y^3)
1 + y^2 = 6y
y^2 - 6y +1 = 0

stuck again...

Answer 6

wtf???? well i would get y^2+y=6 -->\[y^2+y-6=0\]

Answer 7

any ideas how to solve that now?

Answer 8

Hmm, how'd you get "y2+y−6=0"? Doesn't look right to me (?)

Answer 9

your equation was: \[1+\frac{1}{y}=\frac{6}{y^2}\] right? now if you multiplay y^2 you get the equation that i posted.

Answer 10

\[1 +\frac{ 1 }{ y } = \frac{ 6 }{ y^{2} }\]

\[\frac{ y+1 }{ y }=\frac{ 6 }{ y^2 }\]

multiplying both the equation by y
we get
\[y+1= \frac{ 6 }{ y }\]

again multiplying by y we get

\[y^2+y=6\]
\[y^2+y-6=0\]
\[\left( y+3 \right)\left( y-2 \right)\]
therefore y=-3 or y=2

Answer 11

if you multiply by y2 you get,
\[y^{2} + \frac{ y^{2} }{ y } = \frac{ 6 }{ y^{2} } \times y^{2}\]
\[ y^{2} + y - 6 = 0\]

Answer 12

Allright, finally got it! Thanks!

Answer 13

\[y^{2} + 3y - 2y - 6 = 0\]\[(y+3) (y-2) = 0\]y = -3 or 2.

Answer 14

Hmm, I solved y2+y−6=0 using the formula \[x = -(p/2) +- \sqrt((p/2)^2-q)\]

How did you get: y2+3y−2y−6=0
?

Answer 15

he looked for a way to part the function in linear factors. 3y-2y=y so it is your original function

Answer 16

@math0101 dont you know the factorization method? the que you asked is from factorization method ;)

Answer 17

@math0101 Its a method to factorize.

Answer 18

I get this calculation:
y2+3y−2y−6=0
(y+3)(y−2)=0

What I don't get is where you got y2+3y−2y−6=0 from

Answer 19

Ie, what's the step
y2+y−6=0
y2+3y−2y−6=0
?

Answer 20

okay here you will get the idea
http://www.purplemath.com/modules/factquad.htm

Answer 21

thank you!