To solve the given equations simultaneously:

1. Equations:
   • $y = x^2 + 3x - 5$
   • $y = x + 3$

Step 1: Set the equations equal to each other

Since $y = x^2 + 3x - 5$ and $y = x + 3$, equate the two expressions for $y$: $x^2 + 3x - 5 = x + 3$

Step 2: Simplify the equation

Move all terms to one side: $x^2 + 3x - x - 5 - 3 = 0$ $x^2 + 2x - 8 = 0$

Step 3: Solve the quadratic equation

Factorize $x^2 + 2x - 8 = 0$: $x^2 + 4x - 2x - 8 = 0$ $(x + 4)(x - 2) = 0$

Thus, $x = -4$ or $x = 2$.

Step 4: Find the corresponding $y$-values

Substitute each $x$ into $y = x + 3$:

• For $x = -4$: $y = -4 + 3 = -1$
• For $x = 2$: $y = 2 + 3 = 5$

Final Solution:

The solutions are: $(x, y) = (-4, -1) \quad \text{and} \quad (x, y) = (2, 5)$

Would you like me to verify or expand on this solution?

Relative Questions:

1. How do you confirm the solutions graphically?
2. What does the intersection of these two equations represent geometrically?
3. Can this method be used for equations involving higher-degree polynomials?
4. What happens if the equations have no real solutions?
5. How can you solve this system of equations using matrices?

Tip: Always verify your solutions by substituting them back into the original equations.