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
OA: D
inequality
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- ajith
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The options for a are {-6,-5,....0,.....4,5,6}rahul.s 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
OA: D
a cant be odd since if a is odd b cannot be an integer (3a+2b =50)
so a can be {-6,-4, -2,0,2,4,6}
So 7 possible solutions
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- sars72
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i assume the second statement is 7 > mod(a) where mod stands for modulus
so, we have 50 = 3a+2b and -7<a<7
-> 2b = (50-3a) -> b = (50-3a)/ 2
Since b has to be an integer, the result of (50-3a) should be even i.e. divisible by 2
So, that leaves us with the possibile values of a as -6,-4,-2,0,2,4,6
Therefore there are 7 possible values of a and thus 7 pussible values for b
So, the answer is 7
so, we have 50 = 3a+2b and -7<a<7
-> 2b = (50-3a) -> b = (50-3a)/ 2
Since b has to be an integer, the result of (50-3a) should be even i.e. divisible by 2
So, that leaves us with the possibile values of a as -6,-4,-2,0,2,4,6
Therefore there are 7 possible values of a and thus 7 pussible values for b
So, the answer is 7