# Thread: writing a paper on how algebra isn't always boring HELP

1. ## writing a paper on how algebra isn't always boring HELP

Hey all, I am currently writing a paper on how algebra does not always have to be boring and that there are many problems that can be out of the box and challenging. My paper deals with five or so problems and how to solve them. Right now I am suck on two problems. Maybe someone could shed some light on these for me??? lol I am starting to run out of scrap paper.

1. if abc=1 and 1/a+1/b+1/c=a+b+c verify that at least one of a, b, or c is equal to 1

So far I have figured out that in order to solve this either a=1 b=1 or c=1 and either a-1=0 b-1=0 or c-1=0 and then I tried to set it up like (a-1)(b-1)(c-1)= with the entire thing equal to 0. this is how far I have gotten on that problem.

and

2. It takes three days for a boat to travel down a river (i.e. with the current) from point A to point B, and it takes four days for it to come back. How long will it take a log to float from A to B?

for this problem I know that I will be working with the formula for velocity. I also know that I will be using three different velocities, the velocity of the boat, the velocity of the river, and the velocity of the log which is essentially just the velocity of the river because a log travels by current.

Any help with these two problems would be very much appreciated!

2. ## Re: writing a paper on how algebra isn't always boring HELP

problem two doesn't provide enough information to be solved

Let $d$ be the distance from $A$ to $B$

$\dfrac{d}{v_b+v_c}=3$

$\dfrac{d}{v_b-v_c}=4$

$3(v_b+v_c) = 4(v_b-v_c)$

$7v_c = v_b$

$v_c = \dfrac{v_b}{7}$

and that's as far as you can go.

One either needs a hard value for $d$, or a hard value for $v_b$ to solve for $v_c$ and thus the velocity of the log.

3. ## Re: writing a paper on how algebra isn't always boring HELP

Actually, the second problem can be solved.

$t_c =\frac d {v_c} = \frac {3 (v_b+v_c)}{v_c} = \frac {3(8 v_c)}{v_c}$ = 24 days.

4. ## Re: writing a paper on how algebra isn't always boring HELP

$7v_c = v_b \implies v_b+v_c = 8v_c$

traveling at a speed of $8v_c$ downstream takes 3 days to make good the distance from A to B ... traveling at 1/8 that speed (just the current) would take 8 times longer, 24 days.

5. ## Re: writing a paper on how algebra isn't always boring HELP

Originally Posted by ChipB
Actually, the second problem can be solved.

$t_c =\frac d {v_c} = \frac {3 (v_b+v_c)}{v_c} = \frac {3(8 v_c)}{v_c}$ = 24 days.
Where does the 8 come from?

6. ## Re: writing a paper on how algebra isn't always boring HELP

For the first, you can rearrange abc=1 to a = 1/(bc). Then sub that into the other equation:

$bc + \frac 1 b + \frac 1 c = \frac 1 {bc} + b + c$

Multiply through by bc:

$b^2c^2 + c + b = 1 + b^2c + c^2 b$

Do a bit of rearranging:

$b^2c^2 - b^2c -c^2b + b+c-1 = 0$

Now comes the trick: this can be factored. A bit of cheating here: since you know that the value of a, b, or c must be 1, you can let a = 1, which means that that b = 1/c (because abc=1). Hence the above equation must be factorable with one of the terms being (b-1/c). If you do the long division you get:

$(b- \frac 1 c)(bc^2-bc-c^2+c) = 0$

The second part of this equation is also factorable:

$bc^2 - bc -c^2 + c = c(c-1)(b-1)$

Putting it all together:

$(b-\frac 1 c)c(b-1)(c-1) = 0$

We know c does not equal 0, as that would make 1/c undefined. So divide through by c:

$(b-\frac 1 c)(b-1)(c-1) = 0$

Hence possible solutions are b=1/c which means a = 1, and/or b = 1, and/or or c = 1

7. ## Re: writing a paper on how algebra isn't always boring HELP

Originally Posted by njcountrygirl
Where does the 8 come from?
From $v_b = 7 v_2$. Hence $(v_b+v_c) = 7 v_c + v_c = 8 v_c$

8. ## Re: writing a paper on how algebra isn't always boring HELP

Chipb posted his solution to 1 while I was typing the following. This solution is maybe a little easier since it doesn't require any great skill in factoring.