To solve the given system of equations using the addition method, we first write the system of equations:

1. $x^2 + y^2 = 49$
2. $y^2 - 3x = 49$

Step 1: Eliminate one variable

Subtract the second equation from the first to eliminate $y^2$: $(x^2 + y^2) - (y^2 - 3x) = 49 - 49$ $x^2 + y^2 - y^2 + 3x = 0$ $x^2 + 3x = 0$

Step 2: Solve for $x$

Factor the equation: $x(x + 3) = 0$

Thus, $x = 0$ or $x = -3$.

Step 3: Substitute into one of the original equations

We substitute $x = 0$ and $x = -3$ into $x^2 + y^2 = 49$ to solve for $y$.

Case 1: $x = 0$

$0^2 + y^2 = 49$ $y^2 = 49 \implies y = \pm 7$

Thus, the solutions are $(0, 7)$ and $(0, -7)$.

Case 2: $x = -3$

$(-3)^2 + y^2 = 49$ $9 + y^2 = 49$ $y^2 = 40 \implies y = \pm \sqrt{40} = \pm 2\sqrt{10}$

Thus, the solutions are $(-3, 2\sqrt{10})$ and $(-3, -2\sqrt{10})$.

Final Solution

The solution set is: $\{ (0, 7), (0, -7), (-3, 2\sqrt{10}), (-3, -2\sqrt{10}) \}$

---

Do you want further clarification or details?

Here are 5 related questions to expand your understanding:

1. How do we use substitution instead of addition to solve this system?
2. Can we verify these solutions by substituting them back into both equations?
3. What are the graphical interpretations of these solutions?
4. How does factoring help solve quadratic equations in general?
5. Can the elimination method always be used for nonlinear systems?

Tip: Always check your solutions by substituting them back into both original equations to ensure they satisfy the system.