Let's analyze the problem in the image you uploaded. You have a system of two equations:

1. $y = x + 3$
2. $x^2 + y^2 = 5$

a) Show that $x^2 + 3x + 2 = 0$

To do this, we will substitute the expression for $y$ from the first equation into the second equation.

Start with the first equation: $y = x + 3$ Substitute $y = x + 3$ into the second equation: $x^2 + (x + 3)^2 = 5$

Expand $(x + 3)^2$: $x^2 + (x^2 + 6x + 9) = 5$

Simplify the equation: $x^2 + x^2 + 6x + 9 = 5$ $2x^2 + 6x + 9 = 5$ Now subtract 5 from both sides: $2x^2 + 6x + 4 = 0$ Divide the entire equation by 2 to simplify: $x^2 + 3x + 2 = 0$ Thus, we have shown that $x^2 + 3x + 2 = 0$.

b) Solve the simultaneous equations

To solve for $x$ and $y$, we now solve the quadratic equation $x^2 + 3x + 2 = 0$.

Factor the quadratic equation: $(x + 1)(x + 2) = 0$ Set each factor equal to zero: $x + 1 = 0 \quad \text{or} \quad x + 2 = 0$ Solving these gives: $x = -1 \quad \text{or} \quad x = -2$

Now, substitute these values of $x$ into the first equation $y = x + 3$ to find the corresponding values of $y$.

For $x = -1$: $y = -1 + 3 = 2$

For $x = -2$: $y = -2 + 3 = 1$

Thus, the solutions are:

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

Final Answer:

• The solutions are $(-1, 2)$ and $(-2, 1)$.

Would you like more detailed steps or have any questions about this process?

Here are some related questions:

1. How do you solve a quadratic equation using the quadratic formula?
2. What happens if you have a system of equations with no solutions?
3. How do you verify that a pair of values satisfies both equations in a system?
4. What are the properties of a system of equations that can have infinitely many solutions?
5. How do you solve for variables when they appear in both linear and quadratic equations?

Tip: Always check if the quadratic equation can be factored easily before using the quadratic formula!