Solve the system x² + y = 2 and −x² + y = 10.

Step-by-step solution

1. 1

   Spot the elimination that the system is built for. The two equations contain $+x^{2}$ and $-x^{2}$, so simply adding them removes $x$ entirely — no substitution or squaring is needed.

2. 2

   Add the equations.

   $(x^{2}+y)+(-x^{2}+y)=2+10\quad\Longrightarrow\quad 2y=12\quad\Longrightarrow\quad y=6$

3. 3

   Substitute y = 6 back into either equation. Using the first:

   $x^{2}+6=2\quad\Longrightarrow\quad x^{2}=-4$

4. 4

   Interpret the impossible equation. The square of any real number is $\ge 0$, so $x^{2}=-4$ has no real solution. Since $y$ was forced to be $6$, there is no real pair $(x,y)$ satisfying both equations.

   $\boxed{\text{No real solution}}$

5. 5

   Confirm with the second equation and with geometry. Substituting $y=6$ into $-x^{2}+y=10$ gives $-x^{2}=4$, i.e. $x^{2}=-4$ again — the same contradiction, so no algebraic slip occurred. Geometrically, $y=2-x^{2}$ is a downward parabola with vertex $(0,2)$ and $y=10+x^{2}$ is an upward parabola with vertex $(0,10)$; the first never rises above $y=2$ and the second never falls below $y=10$, so their graphs cannot meet.

6. 6

   Note the complex solutions, if the problem allows them. Over the complex numbers $x^{2}=-4$ gives $x=\pm 2i$, so the pairs $(2i,6)$ and $(-2i,6)$ satisfy both equations. Answer choices such as $(4,-14)$, $(-4,-14)$ or $(0,2)$ do not satisfy either equation and can be rejected by direct substitution.