\left\{ \begin{array} { l } { A + B + C + D = 0 } \\ { \frac { A } { 2 } + \frac { B } { 4 } + \frac { C + D } { 2 } = \frac { 1 } { 2 } } \\ { \frac { A } { 3 } + \frac { B } { 9 } + \frac { 2 C + P } { 5 } = \frac { 8 } { 45 } } \\ { \frac { A } { 4 } + \frac { B } { 16 } + \frac { 3 C + D } { 10 } = \frac { 3 } { 40 } } \end{array} \right.
Solve for A, B, C, D, P
A=12-6P
B=-2
C=\frac{9P}{2}-9
D=\frac{3P}{2}-1
P\in \mathrm{R}
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