How do I solve for $g$ in $12g = 12 \left(\frac{2}{3g} - 1\right) + 11$? For this specific problem, I somehow keep coming up with the wrong answer. Can someone help me?
For the problem, I need to solve for $g$.
$$12g = 12 \left(\frac{2}{3g} - 1\right) + 11$$
Here is how I am trying to solve it:
$$12 \cdot \frac{2}{3g} = 8g$$
$$12 \cdot -1 = -12$$
$$12g = 8g - 12 + 11$$
(Then I subtract $8g$ from $12g$ and $8g$)
$4g = -12 + 11$
(Then I add $12$ to $-12$ and $11$)
$$4g = 23$$
(Then I divide $4$ from $4g$ and $23$)
$$g = 5.75$$
So, shouldn't $g$ equal $5.75$? But when I plug it in the equation to check my answer, both answers do not match. Where am I going wrong?

So you have the equation $12g = 12 (\frac{2}{3g} - 1) + 11$. You want to find $g$. $$12g=\frac{24}{3g}-12+11$$ $$\implies 12g=\frac{8}{g}-1$$ $$\implies 12g^2=8-g$$ $$12g^2+g-8=0$$ $$g=\boxed{\frac{-1\pm\sqrt{385}}{24}}$$
However, if you mean $12g=12(\frac{2}{3g-1})+11$, then we have $$12g=\frac{24}{3g-1}+11$$ $$\implies 12g(3g-1)=24+33g-11$$ $$\implies 36g^2-12g=33g+13$$ $$\implies 36g^2-45g-13=0$$ This is a quadratic, so use the quadratic formula getting $$g=\boxed{\frac{15\pm\sqrt{433}}{24}}$$
I’m sure there is some limit of $g$ to eliminate on of the solutions, such as “no negative numbers”, so I think you can take it from there... given that I’m right :P.
