If 13x + < 2z and z + 3 = 5y2 then
Pick one:
A) x is necessarily less than y B) x is necessarily greater than y C) x is necessarily equal to y D) None of the above is necessarily true E) Cannot be determined
Explanation :   =&gt; 13x + 1 &lt; 2z and z + 3 = 5y2   =&gt; 13x + 1 &lt; 2(5y2 - 3).   =&gt; 13x + 1 &lt; 10y2 - 6.   =&gt; 13x + 7&lt; 10y2 , put x = 1   20 &lt; 10y2 =&gt; y2 &gt; 2   y2 &gt; √2 =&gt; (y2 - 2)&gt; 0.   Answer : None of the above is necessarily true
<result>D</result>
If 13x + < 2z and z + 3 = 5y2 then