50 = 3a + 2b
7 > |–a|
If a and b are both integers, how many possible solutions are there to the system above?
Remember that |–a| = |a|
e.g. |–3| = 3 = |3|
If |a| < 7, this implies that the absolute value of a is less than 7. So a could range from -6 to 6 (since a can only be an integer).
Those are 13 values.
Now look at this: 50 = 3a + 2b
50 - 3a = 2b
(50 - 3a)/2 = b
Since b has to be integer too, 50 - 3a must be divisible by 2. Since 50 is even, a should be even too (Even - Even = Even).
From -6 to 6, there are 7 even numbers and 6 odd numbers.
So there are 7 possible solutions (a = -6, b = 34; a = -4, b = 31 etc)
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