CDF in Probability $$f(x)=\left\{\begin{array}{ll}
C & \text { for }-3 \leq x<3 \\
Dx & \text { for } 3 \leq x<5 \\
0 & \text { otherwise }
\end{array}\right.$$
Given that $P(-3 \leq x < 3) = \frac{2}{3}P(3 \leq x < 5)$, What is the cumulative distribution function $CDF(x)$.
So far I have tried to solve the following:
$P(-3 \leq x < 3) = \frac{2}{3}P(3 \leq x < 5)$
$\int_{-3}^x f(x)dx$ = $\frac{2}{3}  \int_{3}^x f(x)dx$
$\int_{-3}^x (C)dx$ = $\frac{2}{3}  \int_{3}^x (Dx)dx$
$\int_{-3}^x (C)dx$ = $ \frac{2}{3}  \int_{3}^x (Dx)dx$
$ \left[Ax\right]_-3^x $ = $\frac{2}{2*3} \left[D* x^{2}\right]_3^x $
C [3 - (x)] = $ \frac{1}{3} \left [D * (x^{2} - 3^{2}\right] $
(3-x)* C = $ \frac{D*(x^{2} - 3^{2})}{3} $
C = $ -\frac{D*(x + 3)}{3} $
How do I find the values for C and D and then calculate the CDF?

$\int_{-3}^{x}f(t)dt \neq \frac{2}{3} \int_{3}^{x}f(t)dt $,   this need not be true
but
$\int_{-3}^{3}f(x)dx = \frac{2}{3} \int_{3}^{5}f(x)dx $
other equation will be
$\int_{-\infty }^{\infty }f(x)dx = 1 $
Get $C,D$ from above two equations
Finally CDF = $\int_{-\infty }^{x }f(t)dt$
