# Trying to solve an equation, but a bit confused with changing of signs

• Feb 18th 2013, 05:38 PM
Odonsky
Trying to solve an equation, but a bit confused with changing of signs
It's been a loooong time since I've worked on math and I am stuck trying to solve an equation from my algebra class.

t / t - 10 = t + 10 / 15

I determined the LCD to be 15(t-10).

Clearing fractions by multiplying both sides of the equation by the LCD, I get:

15(t-10) {t / t - 10} = 15(t-10) {t+10 / 15}, which equals 15t = t^2 - 100.

This is the part where I get confused, regarding changing of signs as a result of writing the equation in standard form. I got this far because I already know the answer:

So if 15t = t^2 - 100, to rewrite in standard form I am instructed to translate the equation to:

t^2 - 15t - 100 = 0. Where does the negative sign between t^2 & 15t come into play?

Factored equation is (t+5) (t-20) = 0.

On a side note, is there a way to write fractions or degrees when posting other than using ' / ' or ^ ?
• Feb 18th 2013, 05:49 PM
HallsofIvy
Re: Trying to solve an equation, but a bit confused with changing of signs
Quote:

Originally Posted by Odonsky
It's been a loooong time since I've worked on math and I am stuck trying to solve an equation from my algebra class.

t / t - 10 = t + 10 / 15

First, you mean "t/(t- 10)= (t+ 10)/15" don't you? What you wrote was "1- 10= t+ (2/3)" because t/t= 1 and 10/15= 2/3.

I determined the LCD to be 15(t-10).

Clearing fractions by multiplying both sides of the equation by the LCD, I get:

15(t-10) {t / t - 10} = 15(t-10) {t+10 / 15}, which equals 15t = t^2 - 100. [/quote]
Yes, this is correct.

Quote:

This is the part where I get confused, regarding changing of signs as a result of writing the equation in standard form. I got this far because I already know the answer:

So if 15t = t^2 - 100, to rewrite in standard form I am instructed to translate the equation to:
You want to get "0" on the right side so subtract 15t from both sides:
16t- 15t= t^2- 100- 15t which is the same as 0= t^2- 15t- 100. Of course, if a= b, then b= a so that is the same as
t^2- 15t- 100= 0.

Quote:

t^2 - 15t - 100 = 0. Where does the negative sign between t^2 & 15t come into play?

Factored equation is (t+5) (t-20) = 0.

On a side note, is there a way to write fractions or degrees when posting other than using ' / ' or ^ ?
• Feb 18th 2013, 06:08 PM
Odonsky
Re: Trying to solve an equation, but a bit confused with changing of signs
Quote:

Originally Posted by HallsofIvy
First, you mean "t/(t- 10)= (t+ 10)/15" don't you? What you wrote was "1- 10= t+ (2/3)" because t/t= 1 and 10/15= 2/3.

Yes, you are correct.

Quote:

Originally Posted by HallsofIvy
You want to get "0" on the right side so subtract 15t from both sides:
16t- 15t= t^2- 100- 15t which is the same as 0= t^2- 15t- 100. Of course, if a= b, then b= a so that is the same as
t^2- 15t- 100= 0.

Ah, I see it now. I wasn't subtracting the 15t originally from on the left side of the equation. Makes perfect sense now.

Thank you very much!
• Feb 18th 2013, 08:45 PM
LanellePalmer
Re: Trying to solve an equation, but a bit confused with changing of signs
I wasn't subtracting the 15t originally from on the left side of the equation. Makes perfect sense now.

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