7-10z>5,
7>5+10z,
7-5>10z,
2>10z,
1>5z is the same as 5z<1, z<⅕ is the solution.
CHECK
Let z=0 then 7-0>5 is true because 0<⅕.
Let z=⅖ then 7-4<5 because 3<5 and ⅖>⅕ which is contrary to z having to be less than ⅕.
When z=⅕, 7-10z=7-2=5 which is not greater than 5 so z cannot be ⅕.