Solve the System of Equations: $x + xy + y = 2 + 3\sqrt{2}$ and $x^2 + y^2 = 6$

Question

Solve the system of equations: $x + xy + y = 2 + 3\sqrt{2}$ and $x^2 + y^2 = 6$

Accepted Answer

$\begin{cases} x = \sqrt{2} + 1 \\ y = \sqrt{2} - 1 \end{cases}$ or $\begin{cases} x = \sqrt{2} - 1 \\ y = \sqrt{2} + 1 \end{cases}$

Answer

Step 1: Recall the algebraic identity $x^2 + y^2 = (x + y)^2 - 2xy$. Substitute $x^2 + y^2 = 6$ into it, we get $(x + y)^2 - 2xy = 6$. Let $m = x + y$ and $n = xy$, so the equation becomes $m^2 - 2n = 6$, which can be rearranged to $n = \frac{m^2 - 6}{2}$. Step 2: Rewrite the first equation $x + xy + y = 2 + 3\sqrt{2}$ as $m + n = 2 + 3\sqrt{2}$. Step 3: Substitute $n = \frac{m^2 - 6}{2}$ into $m + n = 2 + 3\sqrt{2}$, we get $m + \frac{m^2 - 6}{2} = 2 + 3\sqrt{2}$. Multiply both sides by 2 to eliminate the denominator: $2m + m^2 - 6 = 4 + 6\sqrt{2}$. Rearrange into a quadratic equation: $m^2 + 2m - (10 + 6\sqrt{2}) = 0$. Step 4: Solve the quadratic equation for $m$ using the quadratic formula $m = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=2$, $c=-(10+6\sqrt{2})$. Calculate the discriminant: $\Delta = 2^2 - 4\times1\times(-(10+6\sqrt{2})) = 4 + 40 + 24\sqrt{2} = 44 + 24\sqrt{2} = (6 + 2\sqrt{2})^2$. Then $m = \frac{-2\pm(6 + 2\sqrt{2})}{2}$. We take the positive root (since the negative root will lead to no real solutions for $x$ and $y$): $m = \frac{-2 + 6 + 2\sqrt{2}}{2} = 2 + \sqrt{2}$. Step 5: Substitute $m = 2 + \sqrt{2}$ into $n = \frac{m^2 - 6}{2}$. Calculate $m^2 = (2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2}$, so $n = \frac{6 + 4\sqrt{2} - 6}{2} = 2\sqrt{2}$. Step 6: Now we know $x + y = 2 + \sqrt{2}$ and $xy = 2\sqrt{2}$, so $x$ and $y$ are roots of the quadratic equation $t^2 - (2 + \sqrt{2})t + 2\sqrt{2} = 0$. Factor the equation: $(t - (\sqrt{2} + 1))(t - (\sqrt{2} - 1)) = 0$. The roots are $t = \sqrt{2} + 1$ and $t = \sqrt{2} - 1$, which gives the two sets of solutions for $x$ and $y$.