To solve the given system of equations using the elimination method:

The equations are: x^2 - 3y = -3 \tag{1} x^2 + 5y = 29 \tag{2}

Step 1: Subtract equation (1) from equation (2)

$(x^2 + 5y) - (x^2 - 3y) = 29 - (-3)$ Simplify: $x^2 - x^2 + 5y + 3y = 29 + 3$ $8y = 32$ $y = 4$

Step 2: Substitute $y = 4$ into one of the original equations

Using equation (1): $x^2 - 3(4) = -3$ $x^2 - 12 = -3$ $x^2 = 9$ $x = \pm 3$

Step 3: Write the solution as ordered pairs

The solutions are: $(3, 4) \quad \text{and} \quad (-3, 4)$

Final Answer:

$\boxed{(3, 4), (-3, 4)}$

Would you like further details or explanations?

Follow-up Questions:

1. How would the elimination method change if $x^2$ had different coefficients in each equation?
2. What happens if there is no solution for a system of equations like this?
3. Can this method be extended to systems with three variables?
4. How can you verify the solutions using substitution?
5. What alternative methods could solve this system?

Tip:

Always double-check solutions by substituting them back into both original equations to ensure they satisfy the system.