u0422811 wrote:
50 = 3a + 2b
7 > |–a|
If a and b are both integers, how many possible solutions are there to the system above?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Its very simple.Before solve this solution lets discus 2 things:-
1) when we write |-x| this means the same as |x|.
2) Mod is very easy concept if you solve mod question by considering as a distance. when a mod is written as |x-(a)| = b, this means the distance from point 'a' (both side left and right of 'a' on number line) is b. |x-(a)| < b means the distance is between the two extreme distance(left and right side of 'a' on number line, considering the max distance is 'b' from 'a' - as per this scenario.....hence the value of 'a' must be between these two extremes. |x-(a)| > b means the distance is greater than the distance of 'b'..i.e the value of a could be anywhere more than 'b'.
Now come to the question. First its given|-a| < 7 ==> |a| < 7 ===> |a-0| < 7==> the distance from zero is less than 7. So the point will be -7 and 7, as distance from 0 to (-7) is 7 and distance from 0 to 7 is 7. Thus, the value of a will lie in between -7 to 7....i.e the value of a (integer given in ques) would be -6, -5, -4,-3,-2,-1,0,1,2,3,4,5,6.
Now, lets move to equation 3a + 2b = 50 ==> b = 25 -(3/2)a. According to question, b is an integer, hence to make b integer a must be divisible by 2. Now remove the value which can not be divisible by 2 from the possible values of a. It will remain with -6,-4,-2,0,2,4,6...i.e total = 7...
hence ans is D+1 Kudos if it helped you.