Equation rearrangement

nosit

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Aug 9, 2020
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In order to isolate the 1,0725^N, I tried to replicate what I saw in the picture below by rearranging the first equation, but it is not working.

I also tried to see if I could factor something, but it seems not to be possible.

Would someone know how to rearrange it and could show the rearrangement process ?

Thank you in advance.

1618088871004.png
 
The basic form of the equation is Ax + B(x-1) = C, where x represents 1.0725^N. What would you do to solve that?

Alternatively, you might multiply the equation by 0.0725 to eliminate fractions, distribute, and collect terms in the unknown. I'd say that is what they did.
 
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@Dr.Peterson, I am almost there.
Just got stuck to disentangle this part I highlighted below in yellow . Could you kindly help me on it?

1618141867078.png
 
Without wishing to butt in, I would rewrite your first line (as Dr.Peterson suggested) replacing [MATH]1.0725^N[/MATH] with [MATH]x[/MATH]. I would then multiply out the second bracket [MATH]5000(x-1)[/MATH]
 
@Dr.Peterson, I am almost there.
Just got stuck to disentangle this part I highlighted below in yellow . Could you kindly help me on it?

View attachment 26395
You failed to distribute the 5000.

You can't divide by a sum the way you did, dividing each term by a different part of the sum. It is not true that [MATH]\frac{ax+by}{a+b}=x+y[/MATH]. You will need to combine like terms and then divide by the single coefficient.
 
Butting in myself, I extend lex’s post to show you why what he advises is very useful. In fact, the technique is commonplace in calculus. Why it is not taught in algebra is beyond my comprehension.

You have this messy equation

[MATH]10050 * 1.0725^N + 5000 * \dfrac{1}{0.0725} * (1.0725^n - 1) = 60000.[/MATH]
It will take time to manipulate that beast, and the opportunity for error is great. So SUBSTITUTE VARIABLES.

[MATH]\text {Set } x = 1.075^N.[/MATH]
[MATH]\therefore 10050x + \dfrac{5000}{0.0725} (x - 1) = 60000.[/MATH]
Already it looks less strange, looks easier to work with, and looks less error prone.

Frankly I would not proceed as your example does: I would first CLEAR FRACTIONS.

That is simple in this case. Just multiply both sides of the equation by 0.0725 which gives us this:

[MATH]728.625x + 5000(x - 1) = 4350 \implies 728.625x + 5000x - 5000 = 4350 \implies[/MATH]
[MATH]5728.625x = 4350 + 5000 = 9350 \implies x = \dfrac{9350}{5728.625} \approx 1.632.[/MATH]
First year algebra. Distribution error is very unlikely. Now substitute back

[MATH]x \approx 1.632 \implies 1.0725^N \approx 1.632.[/MATH]
You will be told that substituting simple variables for expressions and clearing fractions are unnecessary mathematically. True. But if they make things quicker and less error prone, they can be SUPER helpful.
 
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