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# Hi good people!,

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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 $$-3$$ ??

Nov 10, 2017

#1
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[ 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

Nov 10, 2017
edited by CPhill  Nov 10, 2017
#2
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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
+98102
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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!!!!

Nov 10, 2017
#4
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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
+98102
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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

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

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

Nov 10, 2017
#6
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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
#7
+98102
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Yep......bedtime calls...!!!!!

Nov 10, 2017