$R = 30 + 50 = 80 \, \text{kN}$
$xR = 4(30)$
$x = 120/R$
$x = 120/80$
$x = 1.5 \, \text{m}$
Maximum moment under 30 kN wheel:
$\Sigma M_{R2} = 0$
$8R_1 = 2.75(80)$
$R_1 = 27.5 \, \text{kN}$
$M_{To\,\,the\,\,left\,\,of\,\,30\,\,kN} = 2.75R_1$
$M_{To\,\,the\,\,left\,\,of\,\,30\,\,kN} = 2.75(27.5)$
$M_{To\,\,the\,\,left\,\,of\,\,30\,\,kN} = 75.625 \, \text{kN}\cdot\text{m}$
Maximum moment under 50 kN wheel:
$\Sigma M_{R1} = 0$
$8R_2 = 3.25(80)$
$R_2 = 32.5 \, \text{kN}$
$M_{To\,\,the\,\,right\,\,of\,\,50\,\,kN} = 3.25R_2$
$M_{To\,\,the\,\,right\,\,of\,\,50\,\,kN} = 3.25(32.5)$
$M_{To\,\,the\,\,right\,\,of\,\,50\,\,kN} = 105.625 \, \text{kN}\cdot\text{m}$
Thus, $M_{max} = 105.625 \, \text{kN}\cdot\text{m}$ answer
The maximum shear will occur when the 50 kN is over a support.
$\Sigma M_{R1} = 0$
$8R_2 = 6.5(80)$
$R_2 = 65 \, \text{kN}$
Thus, $V_{max} = 65 \, \text{kN} \,\,$ answer
