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

Sep 18, 2016

$\left(x , y\right)$ is one of:

$\left(1 + \sqrt{6} , - 1 + \sqrt{6}\right)$

$\left(1 - \sqrt{6} , - 1 - \sqrt{6}\right)$

$\left(5 , - 2\right)$

$\left(2 , - 5\right)$

#### Explanation:

Notice that:

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

So we can get a quadratic in $\left(x - y\right)$ by subtracting $4$ times the second equation from the first...

$- 14 = 30 - 4 \cdot 11$

$\textcolor{w h i t e}{- 14} = \left({\left(x + y\right)}^{2} + 3 \left(x - y\right)\right) - 4 \left(x y + 3 \left(x - y\right)\right)$

$\textcolor{w h i t e}{- 14} = \left({\left(x + y\right)}^{2} - 4 x y\right) + \left(3 - 12\right) \left(x - y\right)$

$\textcolor{w h i t e}{- 14} = {\left(x - y\right)}^{2} - 9 \left(x - y\right)$

Add $14$ to both ends to get:

$0 = {\left(x - y\right)}^{2} - 9 \left(x - y\right) + 14$

$\textcolor{w h i t e}{0} = \left(\left(x - y\right) - 2\right) \left(\left(x - y\right) - 7\right)$

So $x - y = 2$ or $x - y = 7$

$\textcolor{w h i t e}{}$
Case $x - y = 2$

From the first given equation, we have:

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

$\textcolor{w h i t e}{30} = {\left(2 y + 2\right)}^{2} + 6 = 4 {\left(y + 1\right)}^{2} + 6$

Subtract $6$ from both ends and transpose to get:

$4 {\left(y + 1\right)}^{2} = 24$

Hence:

${\left(y + 1\right)}^{2} = 6$

So:

$y + 1 = \pm \sqrt{6}$

So

$y = - 1 \pm \sqrt{6}$

with corresponding values of $x$ given by $x = y + 2$

So solutions:

$\left(x , y\right) = \left(1 + \sqrt{6} , - 1 + \sqrt{6}\right)$

$\left(x , y\right) = \left(1 - \sqrt{6} , - 1 - \sqrt{6}\right)$

$\textcolor{w h i t e}{}$
Case $x - y = 7$

From the first given equation, we have:

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

$\textcolor{w h i t e}{30} = {\left(2 y + 7\right)}^{2} + 21$

Subtract $21$ from both ends and transpose to get:

${\left(2 y + 7\right)}^{2} = 9$

Hence:

$2 y + 7 = \pm \sqrt{9} = \pm 3$

So:

$2 y = - 7 + 3 = - 4$

or:

$2 y = - 7 - 3 = - 10$

Hence $y = - 2$ or $y = - 5$

Then we have corresponding values for $x$ using $x = y + 7$

Hence solutions:

$\left(x , y\right) = \left(5 , - 2\right)$

$\left(x , y\right) = \left(2 , - 5\right)$

graph{((x+y)^2+3(x-y)-30)(xy+3(x-y)-11) = 0 [-7.22, 12.78, -6.36, 3.64]}