Here is a method I used to solve this question in the most efficient manner that I haven't seen discussedhere.
Stem:\(A+B+C= Even\)
(1)\(A-B-C=E\)
This becomes\(A=E+B+C\)
Important property: In a term of the form (Even number) + X, the (Even number) plays no role in the Even-Odd nature of the term
In turn, this becomes A=B+C,
Plugging into the equation weget:\(B+C+B+C=E?\) this becomes\(2B+2C=E\)
If A & B are integers, this is sufficient, but A & B can both be .1
(2)\( \frac{A-C}{B} =O \)
...
Stem:\(A+B+C= Even\)
(1)\(A-B-C=E\)
This becomes\(A=E+B+C\)
Important property: In a term of the form (Even number) + X, the (Even number) plays no role in the Even-Odd nature of the term
In turn, this becomes A=B+C,
Plugging into the equation weget:\(B+C+B+C=E?\) this becomes\(2B+2C=E\)
If A & B are integers, this is sufficient, but A & B can both be .1
(2)\( \frac{A-C}{B} =O \)
...
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