Pully `3` and `4` are masses. Hence net force on them should be zero . Therefore we take T tension in the shortest string, then tension in other two string will be `2T` and `4T` .
further , if a is the acceleration of `1` in direction , then from the constraint `a_(2)=-7a_(1)`, acceleration of `2` will be `7a` downward.
Writing the equation , `F _("net") = ma` for the two blocks we have
`4T + 2T + T - 10 = 1 xx a`
or `7T - 10 = a` ...(i)
`20 - T = 2 xx (7a)`
or `20 - T= 14a` ...(ii)
Solving three two equation we get,
`T - 1.62N`
and `a = 1.31m//s^(2)`