В арифметической прогрессии S4=12,S8=40. Найдите a1 и d

Решите задачу:

$S_n=\frac{2a_1+(n-1)d}{2}*n\\\\ \left \{ {{S_4=\frac{2a_1+3d}{2}*4=12} \atop {S_8=\frac{2a_1+7d}{2}*8=40}} \right. \\\\ \left \{ {{2a_1+3d=6} \atop {2a_1+7d=10}} \right. \\\\ \left \{ {{7d-3d=10-6} \atop {2a_1+3d=6}} \right. \\\\ \left \{ {{4d=4} \atop {a_1=\frac{6-3d}{2}}} \right. \\\\ \left \{ {{d=1} \atop {a_1=\frac{6-3}{2}}} \right. \\\\ \left \{ {{d=1} \atop {a_1=\frac{3}{2}}} \right. \\\\$