Find the largest value of t such that
\(\frac{13t^2 - 34t + 12}{3t - 2 } + 5t = 6t - 1.\)
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
+129847 [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
