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Hi good people!,

My son brought home teacher's notes he had to go through, so I tried to explain the stuff. I got to a situation that i simply do not understand. please someone explain this:

\({{t-2} \over {t-1}}={{10} \over {3-t}}-1+{{5} \over {t^2-4t+3}}\)

The second step looks like this, which I understand:

\((t-2)(t-3)=-10(t-1)-(t-1)(t-3)+5\)

This becomes:

\(t^2-3t-2t+6=-10t+10-t^2+3t-t-3+5\)

My problem is: The \(-t^2\) and \(+3t\) I can see, but why is it followed by a \(-t\) and then a \(-3\) ??

juriemagic Nov 10, 2017

#1**+2 **

[ t - 2 ] / [ t - 1 ] = 10 / [ 3-t ] - 1 + 5 / [ t^2 - 4t + 3]

Subtract 10 / [3 - t ] form both sides and factor the last denominator

[ t - 2 ] / [ t - 1] - 10 / [ 3 - t ] = -1 + 5 / [ (t - 1) (t - 3) ]

And we can factor a negative out of - 10 / [ 3 - t] so it becomes + 10 / [ t - 3]

[ t - 2 ] / [ t - 1] + 10 / [ t - 3 ] = -1 + 5 / [ (t - 1) (t - 3) ]

Subtract 5 / [ (t - 1) (t - 3) ] from both sides

[ t - 2 ] / [ t - 1] + 10 / [ t - 3] ] - 5 / [ (t - 1) ( t - 3) ] = -1

Get a common denominator on the left = [ [t - 1 ] [ t - 3]...so we have

( [ ( t - 2) ( t - 3) ] + 10 [ t - 1 ] - 5) / [ (t - 3) ( t - 1) ] = - 1

Simplify the left side

( t^2 - 5t + 6 + 10t - 10 - 5 ) / [ ( t - 1) (t - 3) ] = - 1

( t^2 + 5t - 9 ) / [ ( t - 3) ( t - 1) ] = -1

Multiply both sides by [ ( t - 3) ( t - 1) ]

( t^2 + 5t - 9 ) = -1 [ (t - 3) ( t - 1) ]

(t^2 + 5t - 9) = - [ t^2 - 4t + 3 ]

(t^2 + 5t - 9 ] = -t^2 + 4t - 3 rearrange as

2t^2 + t - 6 = 0 factor as

(2t - 3) ( t + 2) = 0

Set each factor to 0 and solve

2t - 3 = 0 t + 2 = 0

Add 3 to both sides Subtract 2 from both sides

2t = 3 t = -2

Divide both sides by 2

t = 3 / 2

So t = -2 , 3/2

CPhill Nov 10, 2017

#2**+2 **

CPill,

Thank you, I see it's slightly different solved than the teachers method...I'm going to study your method and will come back shortly...THANX!!

juriemagic
Nov 10, 2017

#3**+1 **

The thing about this, juriemagic, is there isn't really * one* way to solve these things.....if we asked ten different people, we might get ten different approaches....heck....I'm not even sure mine is necessarily the best....LOL!!!!

CPhill Nov 10, 2017

#4**+1 **

Well your approach I like....I had a look and saw that you "carried" everything over to the left side in steps, and ended with:

-1 (t - 1)(t - 3) on the right side later on..

then all of that became -t^2 +4t -3..which I perfectly understand...

I was hoping to get a reasoning behind the calculation the teacher had done here;

= -10 (t - 1) - (t - 1)(t - 3) + 5 became -10t + 10 - t^2 + 3t - t - 3 + 5...

I do not understand the rule or law to change the signs like that...

I would have written:

= -10 (t - 1) - (t^2 - 3t - t + 3)..which would then become

= - 10 (t - 1) - t^2 + 3t + t - 3..which is EXACTLY what you also got..how did she get "- t" ?

I can see on the paper she used pencil to indicate she was changing the sign of the second t to a "-" inside the bracket, and the "-1" to a "+1" in this part of the sum: = -10 (t - 1) - (t - 1)(t - 3) + 5...why did she do that?...and how would this give the "-t" in any case?...I followed her calculations from thereon and everything calculates to the same end result...

juriemagic
Nov 10, 2017

#5**0 **

Well...let's see

-10 (t - 1) - (t - 1)(t - 3) + 5

Let's break this up like this

-10 (t - 1) - [(t - 1)(t - 3)] + 5

-10t + [-10*-1] - [ t^2 - 3t - 1t + 3 ] + 5

-10t + [-10*-1] - [ t^2 - 4t + 3 ] + 5

-10t + 10 - t^2 + 4t - 3 + 5

-10t + 4t + 10 - 3 + 5 - t^2

-6t + 12 - t^2

-t^2 - 6t + 12

You teacher made a mistake

-10t + 10 - t^2 + 3t* - t* - 3 + 5 should be -10t + 10 - t^2 + 3t

Which leads to my result of -10t + 10 - t^2 + 4t - 3 + 5

CPhill Nov 10, 2017

#6**+1 **

I guess she must have made a mistake, although everything still works together to the same end result...I can see she must have been explaining a "concept" to someone, as she had underlined the "-10t", "+3t" and "-t"....

anycase, I'm putting this to bed now...CPill, you were a great help!!..thanx a lot!!

juriemagic
Nov 10, 2017