Solution to Problem 243 Statically Indeterminate

$R_1 + R_2 = P$

$R_2 = P - R_1$
 

$\delta_1 = \delta_2 = \delta$

$\left( \dfrac{PL}{AE} \right)_1 = \left( \dfrac{PL}{AE} \right)_2$

$\dfrac{R_1 \, a}{AE} = \dfrac{R_2 \, b}{AE}$

$R_1 \, a = R_2 \, b$
 

$R_1 \, a = (P - R_1)b$

$R_1 \, a = Pb - R_1 \, b$

$R_1 \, (a + b) = Pb$

$R_1 \, L = Pb$

$R_1 = Pb / L$       (okay!)
 

$R_2 = P - Pb/L$

$R_2 = \dfrac{P \, (L -b)}{L}$

$R_2 = Pa / L$       (okay!)