решить систему уравнений х/3 +7у=13, -5х+1/2у=16

Решите задачу:

$\left \{ {{\frac{x}{3}+7y=13 \ | \cdot3} \atop {-5x+\frac{1}{2}y=16 \ | \cdot 2}}$

$\right. \left \{ {{x+21y=39} \atop {-10x+y=32}} \right. \left \{ {{x=39-21y} \atop {-10(39-21y)+y=32}} \right. \left \{ {{x=39-21y} \atop {-390+210y+y=32}} \right.\\\\ \left \{ {{x=39-21y} \atop {211y=32+390}} \right. \left \{ {{x=39-21y} \atop {211y=422}} \right. \left \{ {{x=39-21y} \atop {y=422:211}} \right. \left \{ {{x=39-21\cdot2} \atop {y=2}} \right. \left \{ {{x=39-42} \atop {y=2}} \right.\\\\ \left \{ {{x=-3} \atop {y=2}} \right.$