If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,

How to deal with inequalities involving absolute values? First example shows us the so called "number 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, AND flip 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: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.

-6<=y<=0
To get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.

Note : It is clearly mentioned that both x and y are integers.

If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

solving |y+3|<=3
gives
-6<=Y<=0

further equation 2y-3x+6=0
In order to fiind least value of x we nned to maximize y i,e 0
this gives x =2
And least value of y=-6

So product will be -12
Bunuel @VeritasPrepKarsihma Engr2012
Experts please let me know where I am missing something?

Dear reto

Very well explained! Kudos for that.

Just want to ask 2 questions:

1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?
2. Is the question asking you to find the value of the product xy?

Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution

Best Regards

Japinder

Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3

\(+(y + 3) ≤ 3\)
\(y≤ 0\)

\(-(y + 3) ≤ 3\)
\(y\geq{-6}\)

Solve Eq: 2 2y – 3x + 6 = 0

\(x = \frac{(2y + 6)}{3}\)

as range of y: \(-6=<y<=0\)

we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}

lowest xy = 0

Ans : C

is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12

pls. clear the doubt. thank in advance.

If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?

(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above

A simple algebraic approach.
We have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.
2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2

If we take y = -6; then we have :
- 6 = 3x - 6/ 2
-12 = 3x - 6
-6 = 3x
There fore x = -2.
X * Y = -2*-6 = positive 12 --> not the least value

Now lets take y = 0
0 = 3x - 6/2
3x = -6 => 6 = 3x so x =2
X*Y = 2*0 = 0

Would this approach be right? Expert comment would be welcome!