+1124 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 \(-3\) ??
+129852 [ 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
+1124 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!!
+129852
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!!!!
+1124 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...
+129852 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 + t - 3 + 5
Which leads to my result of -10t + 10 - t^2 + 4t - 3 + 5
+1124 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!!
