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Find the largest value of t such that 

\(\frac{13t^2 - 34t + 12}{3t - 2 } + 5t = 6t - 1.\)

 Mar 6, 2019
edited by Guest  Mar 6, 2019
 #1
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t=97

 Mar 6, 2019
 #4
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THANK YOU SO MUCH! 
U SAVED MY HOMEWORK THX

Guest Mar 6, 2019
 #2
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idc!

 Mar 6, 2019
 #3
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umm...if u did not care u did not have to anwser so rudely 

Guest Mar 6, 2019
 #5
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+1

Solve for t:
5 t + (13 t^2 - 34 t + 12)/(3 t - 2) = 6 t - 1

Bring 5 t + (13 t^2 - 34 t + 12)/(3 t - 2) together using the common denominator 3 t - 2:
(4 (7 t^2 - 11 t + 3))/(3 t - 2) = 6 t - 1

Multiply both sides by 3 t - 2:
4 (7 t^2 - 11 t + 3) = 2 - 3 t + 6 t (3 t - 2)

Expand out terms of the left hand side:
28 t^2 - 44 t + 12 = 2 - 3 t + 6 t (3 t - 2)

Expand out terms of the right hand side:
28 t^2 - 44 t + 12 = 18 t^2 - 15 t + 2

Subtract 18 t^2 - 15 t + 2 from both sides:
10 t^2 - 29 t + 10 = 0

The left hand side factors into a product with two terms:
(2 t - 5) (5 t - 2) = 0

Split into two equations:
2 t - 5 = 0 or 5 t - 2 = 0

Add 5 to both sides:
2 t = 5 or 5 t - 2 = 0

Divide both sides by 2:
t = 5/2 or 5 t - 2 = 0

Add 2 to both sides:
t = 5/2 or 5 t = 2

Divide both sides by 5:
t = 5/2                         or                           t = 2/5

 Mar 6, 2019
 #6
avatar+103120 
+2

[13t^2 - 34t + 12]   / [ 3t - 2]   + 5t  = 6t - 1       subtract 5t from both sides

 

[ 13t^2 - 34t + 12 ] / [ 3t - 2 ]  = t - 1            multiply both sides by  3t - 2

 

13t^2 - 34t + 12  =  (3t - 2) ( t - 1)

 

13t^2 - 34t + 12 = 3 t^2 - 5t + 2      rearrange as

 

10t^2 - 29t + 10  =  0

 

(2t - 5) ( 5t - 2)  = 0

 

Set both factors to 0  and solve for t  and we have that

 

t = 5/2      or     t =  2/5

 

So....t = 5/2 is the max value of t

 

 

cool cool cool

 Mar 6, 2019

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