(1) a+2b+3c=12
(2) 2ab+3ac+6bc=48
Solve (1) for 3c
3c = 12-a-2b
Write (2) as
2ab+a(3c)+2b(3c) = 48
Substitute 12-a-2b for 3c
2ab+a(12-a-2b)+2b(12-a-2b) = 48
2ab+12a-a^2-2ab+24b-2ab-4b^2 = 48
-a^2-2ab+12a-4b^2+24b-48 = 0
a^2+2ab-12a+4b^2-24b+48 = 0
a^2+2ab-12a+4b^2-24b+48 = 0
a^2+(2b-12)a+4b^2-24b+48 = 0
Calculate the discriminant B^2-4AC with A=1, B=2b-12, C=4b^2-24b+48
B^2-4AC =
(2b-12)^2-4(1)(4b^2-24b+48) =
(2b-12)^2-4(4b^2-24b+48) =
4b^2-48b+144-16b^2+96b-192 =
-12b^2+48b-48 =
-12(b^2-4b-4) =
-12(b-2)^2
In order for this discriminant to be non-negative,
it must be 0
-12(b-2)^2 = 0
(b-2)^2 = 0
b = 2
Substitute b=2 in (1)
a+2b+3c = 12
a+2(2)+3c = 12
a+4+3c = 12
a+3c = 8
(3) a = 8-3c
Substitute b=2 in (1)
2ab+3ac+6bc = 48
2a(2)+3ac+6(2)c = 48
(4) 4a+3ac+12c = 48
Substitute 8-3c for a in (4)
4(8-3c)+3(8-3c)c+12c = 48
4(8-3c)+3c(8-3c)+12c = 48
32-12c+24c-9c^2+12c = 48
-9c^2+24c+32 = 48
-9c^2+24c-16 = 0
9c^2-24c+16 = 0
(3c-4)(3c-4) = 0
3c-4 = 0
3c = 4
c = 4/3
(3) a = 8-3c
a = 8-3(4/3)
a = 8-4
a = 4
So a=4, b=2, c=4/3
Answer a+b+c = 4+2+4/3 = 12/3+6/3+4/3 = 22/3
Edwin
