1. We are given the equations:
2a+2b+6=5c
and
{a2−4c,b2−2a,c2−2b}={a−c,b−4c,a+b}
where the elements in each set are distinct.
2. According to the set equality, we have:
(a2−4c)+(b2−2a)+(c2−2b)=(a−c)+(b−4c)+(a+b)
Simplifying the right-hand side:
(a−c)+(b−4c)+(a+b)=a−c+b−4c+a+b=2a+2b−5c
Therefore, we have:
a2+b2+c2−2a−2b−4c=2a+2b−5c
3. Combining this with the given equation 2a+2b+6=5c, we substitute 5c=2a+2b+6 into the equation:
a2+b2+c2−2a−2b−4c=−6
Rearranging terms, we get:
a2+b2+c2−2a−2b−4c+6=0
This can be rewritten as:
(a−1)2+(b−1)2+(c−2)2=0
4. Since the sum of squares is zero, each square must individually be zero:
(a−1)2=0⟹a=1
(b−1)2=0⟹b=1
(c−2)2=0⟹c=2
5. We now verify that the triplet (a,b,c)=(1,1,2) satisfies the original conditions:
- For the equation 2a+2b+6=5c:
2(1)+2(1)+6=5(2)⟹2+2+6=10⟹10=10
This is true.
- For the set equality:
{a2−4c,b2−2a,c2−2b}={12−4(2),12−2(1),22−2(1)}={−7,−1,2}
{a−c,b−4c,a+b}={1−2,1−4(2),1+1}={−1,−7,2}
Both sets are {−7,−1,2}, which are equal.
The final answer is (1,1,2).