Absolute value question

[jfranco23]

2x + y = 12
/y/<=12

For how many ordered pairs (x,y) that 2 solutions of the system above are x and y both integers?

A. 7
B. 10
C. 12
D. 13
E. 14

[cramya]

2x+y=12

x=12-y/2


y has to be even satisfying |y|<=12

12,10,8,6,4,2,0,-2,-4,-6,-8,-10,-12

13 ordered pairs

[jfranco23]

how did you get that conclusion?

[DanaJ]

jfranco23: notice that you have the first condition:
2x + y = 12
Now, 2x is obviously even, and so is 12. This is why you have that:
(even number) + y = (even number), making y even as well.

[sureshbala]

Given 2x+y = 12
and |y|<=12.
Hence -12<=y<=12.
So y can take 25 values from -12 to 12.

We have 2x= 12-y
i.e x = (12-y)/2= 6 - y/2
Hence x will be integer if y is divisible by 2.
From -12 to 12, there are 13 values which are divisible by 2.

Hence the answer is 13