Thread: How to make "y" the subject of a given formula

1. How to make "y" the subject of a given formula

Hi guys,

Make $y$ the subject of the formula: $\dfrac{b}{g}=\sqrt{\dfrac{x+2y^{2}}{x-2y^{2}}}.$

I am having trouble with the above stated question. Please guide me, thanks a lot.

2. Yes, i had "transfer" the square root over to (b/g)^2 = (x+2y^2) / ( (x-2y^2) and then i am stuck at here because i don't know what to do with the "(b/g)^2".

3. Hello, FailInMaths!

Do not state the problem in the title . . .

$\text{Solve for }y\!:\;\;\dfrac{b}{g} \;=\;\sqrt{\dfrac{x+2y^2}{x-2y^2}}$

Square both sides: . $\displaystyle \frac{b^2}{g^2} \;=\;\frac{x+2y^2}{x - 2y^2}$

. . . . . . . . $b^2(x-2y^2) \;=\;g^2(x + 2y^2)$

. . . . . . . . $b^2x - 2b^2y^2 \;=\;g^2x + 2g^2y^2$

. . . . . . $2g^2y^2 + 2b^2y^2 \;=\;b^2x = g^2x$

. . . . . . . $2(g^2+b^2)y^2 \;=\;x(b^2-g^2)$

. . . . . . . . . . . . . . $y^2 \;=\;\dfrac{x(b^2-g^2)}{2(b^2+g^2}$

. . . . . . . . . . . . . . . $y \;=\;\pm\sqrt{\dfrac{x(b^2-g^2)}{2(b^2+g^2)}}$

4. Ok, so you have

$\left(\dfrac{b}{g}\right)^{\!\!2}=\dfrac{x+2y^{2}} {x-2y^{2}}.$

I'd recommend multiplying both sides by the denominator of the RHS. What do you get when you do that?

5. Originally Posted by Soroban
Hello, FailInMaths!

Do not state the problem in the title . . .

Square both sides: . $\displaystyle \frac{b^2}{g^2} \;=\;\frac{x+2y^2}{x - 2y^2}$

. . . . . . . . $b^2(x-2y^2) \;=\;g^2(x + 2y^2)$

. . . . . . . . $b^2x - 2b^2y^2 \;=\;g^2x + 2g^2y^2$

. . . . . . $2g^2y^2 + 2b^2y^2 \;=\;b^2x = g^2x$

. . . . . . . $2(g^2+b^2)y^2 \;=\;x(b^2-g^2)$

. . . . . . . . . . . . . . $y^2 \;=\;\dfrac{x(b^2-g^2)}{2(b^2+g^2}$

. . . . . . . . . . . . . . . $y \;=\;\pm\sqrt{\dfrac{x(b^2-g^2)}{2(b^2+g^2)}}$

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).

Originally Posted by Ackbeet
Ok, so you have

$\left(\dfrac{b}{g}\right)^{\!\!2}=\dfrac{x+2y^{2}} {x-2y^{2}}.$

I'd recommend multiplying both sides by the denominator of the RHS. What do you get when you do that?
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!

6. Originally Posted by FailInMaths
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!