Find the largest value of t such that

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

Guest Mar 6, 2019

edited by
Guest
Mar 6, 2019

#5**+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**

Guest Mar 6, 2019

#6**+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

CPhill Mar 6, 2019