Another approach to this is to divide through by the product on the right-hand side of each equation, so that we get:
1/b + 1/a = 4
1/c + 1/b = 5
1/a + 1/c = 3
Now let A = 1/a, B = 1/b and C = 1/c, so the equations become
B + A = 4 (1)
C + B = 5 (2)
A + C = 3 (3)
Subtract (3) from (2) to get B - A = 2
Add this to (1) to get 2B = 6 so B = 3, so b = 1/3
Put B back into (1) to get A = 1, so a = 1
Put B back into (2) to get C = 2, so c = 1/2
.
a+b=4ab b+c=5bc c+a=3ac
a= b(4a - 1) a = c(3a - 1)
b = a/(4a -1) c = a/(3a -1)
b + c = 5bc
a/(4a -1) + a/(3a -1) = 5a^2/[(4a -1)(3a -1)]
[a(3a -1) + a(4a -1)]/ [(4a -1)(3a -1)] = 5a^2/[(4a -1)(3a -1)]
3a^2 - a + 4a^2 - a = 5a^2
2a^2 - 2a = 0
a^2 - a = 0
So
a(a -1) = 0 0 is trivial....so a = 1
So b = a/[4(a) -1] = 1 / [4(1) -1] = 1/3
And c = a/[3(a) -1] = 1 /[3(1)-1] = 1/2
Check
a + b = 1 + 1/3 = 4/3 = 4ab =4(1)(1/3) = 4/3
b + c = 1/2 + 1/3 = 5/6 =5bc = 5(1/3)(1/2) = 5/6
a + c = 1 + 1/2 = 3/2 = 3ac = 3(1)(1/2) = 3/2
Another approach to this is to divide through by the product on the right-hand side of each equation, so that we get:
1/b + 1/a = 4
1/c + 1/b = 5
1/a + 1/c = 3
Now let A = 1/a, B = 1/b and C = 1/c, so the equations become
B + A = 4 (1)
C + B = 5 (2)
A + C = 3 (3)
Subtract (3) from (2) to get B - A = 2
Add this to (1) to get 2B = 6 so B = 3, so b = 1/3
Put B back into (1) to get A = 1, so a = 1
Put B back into (2) to get C = 2, so c = 1/2
.