SOLUTION: Let x ⧪ y = 2x/y . Suppose (a ⧪ b)⧪c − a ⧪ (b ⧪ c) = a, where a, b and c are nonzero integers. What is the sum of all possible positive values of c? Can I get help

Question 1192910: Let x ⧪ y = 2x/y . Suppose (a ⧪ b)⧪c − a ⧪ (b ⧪ c) = a, where a, b and c are
nonzero integers. What is the sum of all possible positive values of c?
Can I get help on this please? Trying to understand how to solve this? Thank you for your help and time.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Based on the definition of the operator symbol given, we can say
a ⧪ b = 2a/b
b ⧪ c = 2b/c

Furthermore,
(a ⧪ b) ⧪ c − a ⧪ (b ⧪ c) = a
(2a/b) ⧪ c − a ⧪ (2b/c) = a
2(2a/b)/c − 2a/(2b/c) = a
(4a)/(bc)− (2a/1)*(c/2b) = a
(4a)/(bc)− (ac)/b = a
(4a)/(bc)− (ac^2)/(bc) = a
(4a-ac^2)/(bc) = a
a(4-c^2)/(bc) = a
(4-c^2)/(bc) = a/a
(4-c^2)/(bc) = 1

Let's now multiply both sides by bc and get everything to one side.
(4-c^2)/(bc) = 1
4-c^2 = bc*1
4-c^2 = bc
0 = c^2+b*c-4

We have a quadratic where c is the variable. Treat b as a constant for now.

If you were to use the quadratic formula to solve for c, then you should get:
and

Through trial and error, and making a table of values, you should find that:

b = -3 leads to c = 4 in the first equation and c = -1 in the second equation
b = 0 leads to c = 2 in the first equation and c = -2 in the second equation; however we'll ignore this case because a,b,c are nonzero
b = 3 leads to c = 1 in the first equation and c = -4 in the second equation

No other integer values of b will produce integer values for c

The second equation produces nothing but negative c values, which we can ignore.

Focusing on the positive c values means:
b = -3 leads to c = 4
b = 3 leads to c = 1

Add up those c values: 4+1 = 5

Answer: 5

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