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# plz help... this is the only question i cant get. THANK YOU IF U HELP

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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

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umm...if u did not care u did not have to anwser so rudely

Guest Mar 6, 2019
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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

2 t = 5 or 5 t - 2 = 0

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

t = 5/2 or 5 t = 2

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

Mar 6, 2019
#6
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[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   Mar 6, 2019