SOLUTION: The following problem was on my final exam last night, and I cannot figure this out. I am trying to simplify this equation: [3 + b(to power of -1)] over [2 + b(to power of -1)

Question 248679: The following problem was on my final exam last night, and I cannot figure this out. I am trying to simplify this equation:
[3 + b(to power of -1)]
over
[2 + b(to power of -1)]
What I did was turn the b to -1 power into 1 over b. Rewrite equation as
(3 + 1 over b)
over
(2 + 1 over b)
Then I multiplied out to remover the denominator of b in each part of the overall equation. Rewrite equation as
3b+1
over
2b+1
Then i multiplied out to remove the denominator of 2b+1. Rewrite as:
(3b+1) (2b+1) which is also 6b squared + 2b + 3b + 1. Rewrite as:
6b squared + 5b + 1.
Is this correct??
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You were doing great until you decided to multiply by . If this were an equation, then you could do that (to both sides). However, you cannot do that here unless you balance that action out by also dividing by . So you really don't get anywhere in terms of simplification (as it makes things worse). So you're just best off stopping at the point where you got

So or in short,

You can verify your answer with a graph (ie the graphs of the original and the final expressions should be the same).

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
[3 + b(to power of -1)]
over
[2 + b(to power of -1)]
================================
[3 + (1/b)]/[2 + (1/b)]
---
[(3b+1)/b] / [(2b+1)/b]
---
Invert the denominator and multiply:
[(3b+1)/b] * [b/(2b+1)]
---
Cancel the "b" factor which is common to numerator and denominator:
---
= (3b+1)/(2b+1)
=====================
Cheers,
Stan H.
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