6 z − 15 ≥ 4 z + 11 6 z − 4 z ≥ 11 + 15 2 z ≥ 26 ∣ : 2 z ≥ 13 x = [ 13 ; ∞ )
4z+11\\ 2z>26\\ z>13\\\\ 6z-15<4z+11\\ 2z<26\\ z<13"> 6 z − 15 > 4 z + 11 2 z > 26 z > 13 6 z − 15 < 4 z + 11 2 z < 26 z < 13
The inequality 4z + 11"> 6 z − 15 > 4 z + 11 is solvable. By isolating z , we find that the solution is 13"> z > 13 . This means that any value of z greater than 13 will satisfy the inequality.
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