# Question #30c88

Oct 24, 2016

$\left(x , y\right) \in \left\{\begin{matrix}0 & 0 \\ - 3 & 2\end{matrix}\right\}$

#### Explanation:

$\left\{\begin{matrix}x + 2 y = 1 \\ {y}^{2} = 1 - x\end{matrix}\right.$

In the second equation, add $x - {y}^{2}$ to both sides to isolate $x$:

$x = 1 - {y}^{2}$

Substitute this value for $x$ into the first equation.

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

Gather the terms on the left hand side.

$- {y}^{2} + 2 y + 1 - 1 = 0$

Combine like terms and multiply by $- 1$ to get a coefficient of $1$ for ${y}^{2}$.

${y}^{2} - 2 y = 0$

This can be factored easily, so we will solve by factoring

$y \left(y - 2\right) = 0$

$\implies y = 0 \mathmr{and} y - 2 = 0$

$\implies y = 0 \mathmr{and} y = 2$

Substitute the possible solutions for $y$ back into $x = 1 - {y}^{2}$ to find the $x$ that goes with them.

$y = 0 \implies x = 1 - {0}^{2} = 1$

$y = 2 \implies x = 1 - {2}^{2} = - 3$

Thus, our potential solutions are

$\left(x , y\right) = \left(1 , 0\right) \mathmr{and} \left(x , y\right) = \left(- 3 , 2\right)$

Checking these answers by substituting them into the original equations, we find they both satisfy the system. Thus, our final result is:

$\left(x , y\right) \in \left\{\begin{matrix}1 & 0 \\ - 3 & 2\end{matrix}\right\}$