Question

Can anyone isolate y?

2y(x^2)+7x(y^2)=408

Answer 1

yes have u tried distributing?

Answer 2

i tried but i cant seem to get y to stay on one side

Answer 3

i got this\[2xy+7y^2=\frac{ 408 }{ x }\]

Answer 4

2y(x^2)+7x(y^2)=408
7x(y^2)+2(x^2)y+7x(y^2)-408=0
Quadratic formula
\[y=\frac{-(2x^2)\pm\sqrt{(2x^2)^2-4(7x)(-408)}}{2(7x)}\]

Answer 5

oh crap i completely forgot about the quadratic formula dang it thank you

Answer 6

Second line should be
7x(y^2)+2(x^2)y+-408=0

Answer 7

ok thank you

Answer 8

Welcome :)

Answer 9

Not sure if it is the right approach though.

Answer 10

Firstly, there is no way of "isolating" y here. You can solve for 'y' given some value of 'x' using the quadratic formula like @Callisto suggested.

@Nogarenshi

Answer 11

well its the only one i got so i will go for it this is for multivariable calculus so i forgot about the quadratic formula

Answer 12

ok thank you

Answer 13

um i am looking at this again and something doesnt seem right can anyone else clarify.

Answer 14

What doesn't seem right?

Answer 15

first open up the bracket: 2y(x^2)+7x(y^2)=408
by opening up you'll have something similar to this 2yx^2+7xy^2=408
then factor out y:. y(2x^2+7x^2)=408
finally divide both side by (2x^2+7x^2)
and your answer will be Y=408/2x^2+7x^2

Answer 16

@hassannoor0480 That's not how it works. By factoring out 'y', there is still a 'y' left with the "7xy". Like I mentioned, there is no way of isolating y here.

Answer 17

ok actually nevermind what castillo said works i notice i wrote something down wrong

Answer 18

Yeah, what Callisto says is true, he isn't factoring anything out:\[

~~~~~\color{red}{(7x)}\cdot y^2+~\color{green}{(2x^2)}\cdot y ~~\color{blue}{- 408} = 0 \\
a = \color{red}{7x},~~~~~~b=\color{green}{2x^2},~c=\color{blue}{-408}\\

\]

Answer 19

yeah genius am wrong i thought y is not squared