If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value T of the product xy?  Pick:	A) -12	B) -3	C) 0	D) 2	E) None of the above
How to deal with inequalities involving absolute values? First example shows us the so callednumber case   In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:   y + 3 ≤ 3   y ≤ 0   So y is 0 or negative   Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality,ANDflip the sign:   y + 3 &gt;= -3   y &gt;= -6   Therefore we have a possible range for y: -6=&lt;y&lt;=0   Ok, so far so good, we're half way through. What about x?   Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.   You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value T.   Hence, Answer 0hoice 0 is the one to go.
<result>C</result>
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value T of the product xy?