Solve the system x² + y² + xy = 7 and x + y + xy = 5.

Step-by-step solution

1. 1

   Introduce the elementary symmetric quantities. Both equations are symmetric in $x$ and $y$, so set

   $s=x+y,\qquad p=xy$

   Using $x^{2}+y^{2}=s^{2}-2p$, the first equation becomes

   $s^{2}-2p+p=s^{2}-p=7$

   and the second is simply $s+p=5$.

2. 2

   Solve the small system in s and p. From the second, $p=5-s$; substituting into the first:

   $s^{2}-(5-s)=7\quad\Longrightarrow\quad s^{2}+s-12=0\quad\Longrightarrow\quad (s+4)(s-3)=0$

   so $s=3$ or $s=-4$.

3. 3

   Test the branch s = 3. Then $p=5-3=2$, and $x,y$ are the roots of

   $t^{2}-st+p=t^{2}-3t+2=0\quad\Longrightarrow\quad (t-1)(t-2)=0$

   giving $\{x,y\}=\{1,2\}$.

4. 4

   Test the branch s = −4. Then $p=5-(-4)=9$, and $x,y$ would be the roots of

   $t^{2}+4t+9=0$

   whose discriminant is $16-36=-20<0$. This branch yields no real solutions and must be discarded — checking $s^{2}\ge 4p$ is the standard test for whether a symmetric pair is realisable over the reals.

5. 5

   State the solutions.

   $\boxed{(x,y)=(1,2)\ \text{or}\ (2,1)}$

6. 6

   Verify. For $(1,2)$: $1+4+2=7$ ✓ and $1+2+2=5$ ✓. The pair $(2,1)$ works identically because the system is symmetric — which also explains why the solutions come in swapped pairs rather than singly.