Equation system with three unknown variables

[Anna55]

Please solve the equationsystem:
ab+ac=2(a+b+c)
ac+bc=4(a+b+c)
bc+ab=8(a+b+c)

First I added all the equations and now I do not know how to continue. Can you please help me? Thank you.

 

[Denis]

ab+ac=2(a+b+c)
ac+bc=4(a+b+c)
bc+ab=8(a+b+c)

You didn't notice that a+b+c = (ab + ac)/2 = (ac + bc)/4 = (ab + bc)/8 ?

 

[soroban]

Hello, Anna55!

Solve the system:

. . ab + ac .= .2(a+b+c) . [1]
. . ac + bc .= .4(a+b+c) . [2]
. . bc + ab .= .8(a+b+c) . [3]

We see that a = b = c = 0 is a solution.


From [1] and [2], we see that: .ac + bc .= .2(ab + ac) . ---> . 2ab - bc + ac .= .0 . [4]

From, [2] and [3], we see that: .bc + ab .= .2(ac + bc) . ---> . ab - bc - 2ac .= .0 . [5]

Subtract [4] - [5]: . ab + 3ac .= .0 . ---> . a(b + 3c) ,.= .0 . ---> . b .= .-3c . [6]

Substitute into [4]: . 2a(-3c) - (-3c)c + ac .= .0 . ---> . 3c^2 - 5ac .= .0 . ---> . c(3c - 5a) .= .0 . ---> . a .= .3c/5 . [7]

Substitute into [1]: .(3c/5)(-3c) + (3c/5)(c) .= .2(3c/5 - 3c + c) . ---> . (2c/5)(3c - 7) .= .0 . ---> . c = 7/3

Substitute into [7]: .a .= .(3/5)(7/3) . ---> . a = 7/5

Substitute into [6]: .b .= .-3(7/3) . ---> . b = -7


Therefore: .a = 7/5, . b = -7, . c = 7/3

 

[tkhunny]

Did you find BOTH solutions?

 

[Denis]

Testing 1 2 3 ....

 

[Anna55]

Thank you Denis and soroban! I understand how to solve it now.