The system of equations presented in the image is:

1. $x^2 + y^2 = 5$
2. $x + y = 1$

To solve this system efficiently, we can proceed by substitution.

Step-by-step approach using substitution:

1. From the second equation, solve for $y$: $y = 1 - x$

2. Substitute $y = 1 - x$ into the first equation: $x^2 + (1 - x)^2 = 5$ Now, expand the square: $x^2 + (1 - 2x + x^2) = 5$ Simplifying this equation: $2x^2 - 2x + 1 = 5$

3. Move all terms to one side: $2x^2 - 2x + 1 - 5 = 0$ $2x^2 - 2x - 4 = 0$

4. Divide through by 2 to simplify: $x^2 - x - 2 = 0$

5. Factor the quadratic equation: $(x - 2)(x + 1) = 0$ So, $x = 2$ or $x = -1$.

6. Substitute each value of $x$ back into $y = 1 - x$:

   • If $x = 2$, then $y = 1 - 2 = -1$.
   • If $x = -1$, then $y = 1 - (-1) = 2$.

Final solutions:

• $(x, y) = (2, -1)$
• $(x, y) = (-1, 2)$

These are the two pairs of solutions.

Is this method efficient?

Yes, substitution is an efficient method in this case because the second equation is simple enough to solve for one variable directly and substitute into the first equation. This avoids more complex techniques like elimination or using matrices.

Do you have any other questions or would you like further details on this method?

Expanding questions:

1. How would this method change if the second equation were more complex?
2. Can this system of equations be solved graphically? How would the solution look?
3. What are alternative methods (like elimination or matrices) to solve this system?
4. How does the solution change if the constant in the first equation changes (e.g., $x^2 + y^2 = 10$)?
5. Can we use the quadratic formula to solve this instead of factoring?

Tip:

For systems involving quadratics, substitution often works best when one equation is linear, as you can easily isolate one variable and substitute into the other equation.