Find the largest value of $c$ such that $\frac{c^2 + 6c -27}{c-3} +2c = 8$.
First, we notice that $\frac{c^2 + 6c - 27}{c - 3}$ can be rewritten as $c+\frac{18}{c-3}$. Thus, our equation becomes $c+\frac{18}{c-3}+2c=8$. We simplify and combine like terms to find that $c+\frac{18}{c-3}+2c=3c+\frac{18}{c-3}=8$. Hence $3c^2-9c+18=8c-24$, which rearranges to $3c^2-17c+18=0$. By the quadratic formula, we find that $c=\frac{17 \pm \sqrt{17^2 - 4(3)(18)}}{6} = \frac{17 \pm \sqrt{13}}{6}$. Since $c$ must be positive, the maximum value is $\frac{17 + \sqrt{13}}{6} = \boxed{\frac{17 + \sqrt{13}}{6}}$.
To find the largest value of $c$ that satisfies the equation $\frac{c^2 + 6c - 27}{c-3} +2c = 8$, we need to solve the equation and determine the range of possible values for $c$.
1. Begin by multiplying both sides of the equation by $c - 3$ to eliminate the fraction:
$(c^2 + 6c - 27) + 2c(c-3) = 8(c - 3)$
Expanding the equation yields:
$c^2 + 6c - 27 + 2c^2 - 6c = 8c - 24$
2. Combine like terms:
$3c^2 - 9 = 8c - 24$
3. Rearrange the equation by moving all terms to one side, setting it equal to zero:
$3c^2 - 8c + 15 = 0$
4. Factor the quadratic:
$(3c - 5)(c - 3) = 0$
5. Set each factor equal to zero and solve for $c$. This gives us two possible solutions:
$3c - 5 = 0 \implies c = \frac{5}{3}$
$c - 3 = 0 \implies c = 3$
6. Verify if the solutions satisfy the original equation:
For $c = \frac{5}{3}$:
$\frac{\left(\frac{5}{3}\right)^2 + 6\left(\frac{5}{3}\right) - 27}{\frac{5}{3} - 3} + 2\left(\frac{5}{3}\right) = 8$
After simplification, we find that the left-hand side of the equation is equal to $8$, so this solution is valid.
For $c = 3$:
$\frac{3^2 + 6(3) - 27}{3 - 3} + 2(3) = 8$
The denominator of the fraction is equal to zero, which means this solution is not valid.
7. Therefore, the largest value of $c$ that satisfies the equation is $\frac{5}{3}$.