6 x 2/2 = 6, because 2/2 is 1 and 6x1 is 6
5^2 +12^2= diagonal^2
25+144=d^2
169= d^2
D= 13
6>4-2x<4
so therefor (4x-2) must be less than 6 and less than 4
it must be
6>4-2x and 4<4-2x
find the intersection
6>4-2x
add 2x to both sides
6+2x>4
subtract 6
2x>-2
divide by 2
x>-1
4>4-2x
add 2x
4+2x>4
subtract 4
2x>0
x>0
so we have
x>0
and x>-1
the range of x>-1 includes most of x>0 so the answer is
x>-1
20/2-10/5. You divide 20 by 2 and subtract that from 10 divided by 5
