If (x^2 + 1)y=6, which of the following is not a possible value for y?  Choices:
A) 3/2 B) 2 C) 1/3 D) 2/13 E) 13/2
Its simple   equation is (x^2 +1)y = 6 ===&gt; x^2 = (6/y) - 1 . As , in LHS, it is x^2 the value has to be positive or zero in RHS.   Hence, (6/y) - 1&gt;= 0 ===&gt; y =&lt; 6. Now only option E has greater value of 'y' than 6.   Thus, correct ans is 13/2
<result>E</result>
If (x^2 + 1)y=6, which of the following is not a possible value for y?