Hi guys,
Make $\displaystyle y$ the subject of the formula: $\displaystyle \dfrac{b}{g}=\sqrt{\dfrac{x+2y^{2}}{x-2y^{2}}}.$
I am having trouble with the above stated question. Please guide me, thanks a lot.
Hi guys,
Make $\displaystyle y$ the subject of the formula: $\displaystyle \dfrac{b}{g}=\sqrt{\dfrac{x+2y^{2}}{x-2y^{2}}}.$
I am having trouble with the above stated question. Please guide me, thanks a lot.
Hello, FailInMaths!
Do not state the problem in the title . . .
$\displaystyle \text{Solve for }y\!:\;\;\dfrac{b}{g} \;=\;\sqrt{\dfrac{x+2y^2}{x-2y^2}} $
Square both sides: .$\displaystyle \displaystyle \frac{b^2}{g^2} \;=\;\frac{x+2y^2}{x - 2y^2}$
. . . . . . . .$\displaystyle b^2(x-2y^2) \;=\;g^2(x + 2y^2)$
. . . . . . . .$\displaystyle b^2x - 2b^2y^2 \;=\;g^2x + 2g^2y^2$
. . . . . . $\displaystyle 2g^2y^2 + 2b^2y^2 \;=\;b^2x = g^2x$
. . . . . . . $\displaystyle 2(g^2+b^2)y^2 \;=\;x(b^2-g^2) $
. . . . . . . . . . . . . . $\displaystyle y^2 \;=\;\dfrac{x(b^2-g^2)}{2(b^2+g^2} $
. . . . . . . . . . . . . . .$\displaystyle y \;=\;\pm\sqrt{\dfrac{x(b^2-g^2)}{2(b^2+g^2)}} $
I was told to specify the question by one of the moderators in the past.
Anyway thanks for the answer, I should change (b/g)^2 = (b^2)/(g^2).
Thanks a lot for all the hints but i found out my problem already. It is to change (b/g)^2 = (b^2)/(g^2).
Thank you everyone!
-Thread Closed-
Glad you've figured out your problem. You do need to state the entire problem in the body of the post, but there's no rule against partially stating the problem in the title of the thread. The thread title should be at least one level more specific than the forum in which you post the thread. Saying, for example, "Algebra Problem", while posting in this subforum, would likely earn you an infraction for useless post title. On the other hand, "Solve for y" would be perfectly acceptable. At least, this is my current understanding of the rules.