\(\frac{13t^2 - 34t + 12}{3t - 2 } + 5t = 6t - 1\)
Find the largest value of t
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\(\frac{13t^2 - 34t + 12}{3t - 2 } + 5t = 6t - 1\\ \frac{13t^2 - 34t + 12}{3t - 2 } = t - 1\\ 13t^2 - 34t + 12=(t-1)(3t-2)\)
1) Now expand the right hand side
2) Take all the terms to the left hand side so the right hand side is 0
3) Solve using the quadratic formula (or any other method)
4) The answer you want is the biggest one.
Now I will leave the algebra for you.
The answer is quite simple, you can check it by plugging it into the original formula and seeing that the right hand side=left hand side.
If you get in a mess you can ask for help but be specific about what your problem is.
Have fun :)
Please no one answer over the top of me. The detail in this answer is perfect as it is.
