\left\{ \begin{array} { l } { \frac { 5 } { 12 } ( 4 + r ) = E } \\ { \frac { 5 } { 54 } ( p + \gamma ) = E } \end{array} \right.
Solve for r, E, p, γ
r=\frac{2\left(p+\gamma -18\right)}{9}
E=\frac{5\left(p+\gamma \right)}{54}
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