Solving an equation with the same variable twice

This may be kind of obvious, but our finance lecturer just assumes we know how to do stuff like this and the slides give no explanation. Jumping from question to answer

How do I get $r$ from this?

$27.397= \displaystyle \frac{1- \displaystyle \frac{1}{(1+r)^{36}} }{r}$

$r=1.558\%$

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This is not an inequality. It's an equation. Confirm it because none of the solutions is 1.558%. – Américo Tavares May 18 '11 at 22:15
John, it looks like you have the inverse of the equation used to determine the monthly payment of a loan without the principal term. $$p = A_0 \frac{r (1 + r)^n}{(1+r)^n - 1}$$ Which arises from solving the recurrence relation $A_{n+1} = (1+r)A_{n} - p$. So I'm assuming you mean $$x = \frac{1 - \frac{1}{(1+r)^n}}{r}$$ with $x = 27.397$ and $n = 36$ – GEL May 18 '11 at 22:38
You won't find a solution algebraically. You need numerics, but the equation is well-behaved. Excel has a built in goal seek that will do this, as will Wolfram Alpha. – Ross Millikan May 18 '11 at 22:41
Your finance lecturer is paid to teach you stuff. Why not ask her how she expects you to solve such equations? – Gerry Myerson May 19 '11 at 0:33

You want to find the root $r>0$ of $$f(r) = \bigg[1 - \frac{1}{{(1 + r)^{36} }}\bigg]\bigg/r - 27.397,$$ that is the $r > 0$ for which $f(r)=0$. Using Wims Function Calculator, I get $r \approx 0.0155764414039$, which is what you want.