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

Sep 20, 2016

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

#### Explanation:

Given:

$\left\{\begin{matrix}{\left(x + y\right)}^{\frac{1}{2}} + 3 \left(x - y\right) = 30 \\ x y + 3 \left(x - y\right) = 11\end{matrix}\right.$

We can isolate the terms in $\left(x + y\right)$ and $x y$ on the left hand side to get:

$\left\{\begin{matrix}{\left(x + y\right)}^{\frac{1}{2}} = 3 \left(10 - \left(x - y\right)\right) \\ x y = 11 - 3 \left(x - y\right)\end{matrix}\right.$

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

$\left\{\begin{matrix}{\left(x + y\right)}^{2} = 81 {\left(10 - \left(x - y\right)\right)}^{4} \\ 4 x y = 44 - 12 \left(x - y\right)\end{matrix}\right.$

Using the identity:

${\left(x + y\right)}^{2} - 4 x y = {\left(x - y\right)}^{2}$

subtract the second equation from the first to get:

${\left(x - y\right)}^{2} = 81 {\left(10 - \left(x - y\right)\right)}^{4} + 12 \left(x - y\right) - 44$

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

${\left(t + 10\right)}^{2} = 81 {t}^{4} + 12 \left(t + 10\right) - 44$

which expands to:

${t}^{2} + 20 t + 100 = 81 {t}^{4} + 12 t + 120 - 44$

Hence:

$81 {t}^{4} - {t}^{2} - 8 t - 24 = 0$

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:

${t}_{1} \approx - 0.69521$

${t}_{2} \approx 0.78593$

For each of these, look at:

$x y + 3 \left(x - y\right) = 11$

Using $x - y = {t}_{n} + 10$, we have:

$x \left(x - \left({t}_{n} + 10\right)\right) + 3 \left({t}_{n} + 10\right) = 11$

Hence:

${x}^{2} - \left({t}_{n} + 10\right) x + \left(3 {t}_{n} + 19\right) = 0$

Then you can use the quadratic formula to find possible values for $x$ and hence for $y = x - \left({t}_{n} + 10\right)$

These $\left(x , y\right)$ pairs then need to be checked against the equation:

${\left(x + y\right)}^{\frac{1}{2}} + 3 \left(x - y\right) = 30$

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:

$\left(x , y\right) \approx \left(6.82734 , - 2.47744\right)$

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Conclusion

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