Problem: if x and y are integers such that | y + 3 | ≤ 3 and 2 y – 3 x + 6 = 0 , what is the least possible value t of the product xy ?
Rationale: "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 > = - 3 y > = - 6 therefore we have a possible range for y : - 6 = < y < = 0 ok , so far so good , we ' re half way through . what about x ? here ' s the formula : 2 y – 3 x + 6 = 0 , rewrite it as 2 y + 6 = 3 x . you can say that 2 y + 6 is a multiple of 3 ( = 3 x ) . 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 choice c is the one to go ."
annotated_formula: multiply(divide(add(6, multiply(negate(add(3, 3)), 2)), 3), negate(add(3, 3)))