We are given the system of equations:

1. $y^2 = 1 - x$
2. $x + 2y = 1$

Step 1: Solve for $x$ in the second equation

From equation (2): $x = 1 - 2y$

Step 2: Substitute $x = 1 - 2y$ into the first equation

Now, substitute this expression for $x$ into equation (1): $y^2 = 1 - (1 - 2y)$ Simplifying the right-hand side: $y^2 = 1 - 1 + 2y$ $y^2 = 2y$

Step 3: Rearrange and solve for $y$

Rearrange the equation to get a quadratic form: $y^2 - 2y = 0$ Factor the equation: $y(y - 2) = 0$ So, the solutions for $y$ are: $y = 0 \quad \text{or} \quad y = 2$

Step 4: Find corresponding $x$ values

• If $y = 0$, substitute into $x = 1 - 2y$: $x = 1 - 2(0) = 1$ So, one solution is $(x, y) = (1, 0)$.

• If $y = 2$, substitute into $x = 1 - 2y$: $x = 1 - 2(2) = 1 - 4 = -3$ So, another solution is $(x, y) = (-3, 2)$.

Final Answer:

The system has two solutions: $(x, y) = (1, 0) \quad \text{and} \quad (x, y) = (-3, 2)$

Do you want more details on any step, or do you have any questions?

Here are 5 related questions to explore:

1. How can you solve a system of nonlinear equations graphically?
2. What other methods can be used to solve nonlinear systems?
3. How does factoring help in solving quadratic equations?
4. What are the properties of solutions to quadratic equations?
5. How do you verify the solutions of a system of equations?

Tip: When solving systems of equations, substitution is especially useful if one equation is already solved for a variable!