My first equations is this: $ d_2 = d - 30.$
My second equations is this: ${1\over d_2 }= {1\over12} - {1\over1.066(d-30)}$
I am trying to solve for $d_2$ in the second equation and then set the two equations equal to each other to solve for d.
I am having trouble solving for $d_2$ in the second equation. I think the fractions are giving me a hard time. Any help is much appreciated, thanks!
On
First, I hate decimals, so let's note that 1.066 equals $\frac{533}{500}$.
Then, since we were essentially given what $d_2$ equals, let's substitute that into our second equation like so:
$$ \frac{1}{d-30} = \frac{1}{12} - \frac{1}{\frac{533}{500}(d-30)}$$ $$ \frac{1}{d-30} = \frac{1}{12} - \frac{500}{533(d-30)}$$
Now we want to multiply through by the LCM (which in our case is simply $12 \cdot 533(d-3)$) $6396(d-30)$ to obtain:
$$ 6396 = 533(d-30) - 6000 $$
Now we have a nice linear equation which we may solve for $d$:
$$ 533d - 21990 = 6396 $$ $$ d = \frac{28386}{533} $$
With $d$ obtained we can easily find $d_2$:
$$ d_2 = d - 30$$ $$ d_2 = \frac{28386}{533} - 30 $$ $$ d_2 = \frac{12396}{533} $$
And we're done.
$$\frac{1}{d_{2}}=\frac{1}{12}-\frac{1}{1.066d_{2}} $$
$$(1+\frac{1}{1.066})\frac{1}{d_{2}}=\frac{1}{12}$$
$$d_{2}=12(1+\frac{1}{1.066})$$
$$d=d_{2}+30$$
$$d=12(1+\frac{1}{1.066})+30$$