We can only use two at a time for the time being.
-6a - 1 ≥ 3a - 10
add 10
-6a + 9 ≥ 3a
add 6a
9 > 9a
divide
1 ≥ a
Now we know a is less than 1 and can fill all a's with 1 in order to find the value of z.
-2z + 3 ≥ 6 * 1 - 1
add 1
-2z + 4 ≥ 6
Subtract 4
-2z ≥ 2
Divide by -2
z ≥ 1
z is ≥ 1, and a is ≤ 1