Question

How do you solve the system of equations: `(x+y)^(1/2)+3(x-y)=30` and `xy+3(x-y) = 11` ?

Answer 1

If we try to solve this algebraically then this one (unavoidably) turns into a quartic with rather messy solutions...

Explanation:

Given:

`{ ((x+y)^(1/2)+3(x-y)=30), (xy+3(x-y) = 11) :}`

We can isolate the terms in `(x+y)` and `xy` on the left hand side to get:

`{ ((x+y)^(1/2) = 3(10-(x-y))), (xy = 11 - 3(x-y)) :}`

Raising the first equation to the `4`th power and multiplying the second by `4` we get:

`{ ((x+y)^2 = 81(10-(x-y))^4), (4xy = 44 - 12(x-y)) :}`

Using the identity:

`(x+y)^2-4xy = (x-y)^2`

subtract the second equation from the first to get:

`(x-y)^2 = 81(10-(x-y))^4+12(x-y)-44`

Letting `t = x-y-10`, we have:

`(t+10)^2 = 81t^4+12(t+10)-44`

which expands to:

`t^2+20t+100 = 81t^4+12t+120-44`

Hence:

`81t^4-t^2-8t-24 = 0`

This quartic has two irrational Real roots and two non-Real Complex roots. The solution is way too messy to present here.

The Real roots are approximately:

`t_1 ~~ -0.69521`

`t_2 ~~ 0.78593`

For each of these, look at:

`xy + 3(x-y) = 11`

Using `x-y = t_n + 10`, we have:

`x(x-(t_n+10)) + 3(t_n+10) = 11`

Hence:

`x^2-(t_n+10)x+(3t_n+19) = 0`

Then you can use the quadratic formula to find possible values for `x` and hence for `y = x-(t_n+10)`

These `(x, y)` pairs then need to be checked against the equation:

`(x+y)^(1/2) + 3(x-y) = 30`

since we raised both sides of this equation to the `4`th power and thus introduced a spurious solution.

The valid solution of the original equation is approximately:

`(x, y) ~~ (6.82734, -2.47744)`

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Conclusion

Attempting to solve this problem algebraically gets too messy. A numerical approach would probably be better.