Question 1192910
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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:
{{{c = (-b+sqrt(b^2+16))/2}}} and {{{c = (-b-sqrt(b^2+16))/2}}}


Through trial and error, and making a table of values, you should find that:
<ul><li>b = -3 leads to c = 4 in the first equation and c = -1 in the second equation</li><li>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</li><li>b = 3 leads to c = 1 in the first equation and c = -4 in the second equation</li></ul>
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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