We need to solve the given simultaneous equations:

1. $x - y = 3$
2. $xy - 5x + y = -13$

Step 1: Express $y$ in terms of $x$ from the first equation

From $x - y = 3$, we get: $y = x - 3$

Step 2: Substitute $y = x - 3$ into the second equation

Substituting $y = x - 3$ into $xy - 5x + y = -13$: $x(x - 3) - 5x + (x - 3) = -13$ Simplify: $x^2 - 3x - 5x + x - 3 = -13$ $x^2 - 7x + x - 3 = -13$ $x^2 - 6x - 3 + 13 = 0$ $x^2 - 6x + 10 = 0$

Step 3: Solve $x^2 - 6x + 10 = 0$ using the quadratic formula

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -6$, and $c = 10$. Substituting these values: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)}$ $x = \frac{6 \pm \sqrt{36 - 40}}{2}$ $x = \frac{6 \pm \sqrt{-4}}{2}$ $x = \frac{6 \pm 2i}{2}$ $x = 3 \pm i$

Step 4: Find $y$ using $y = x - 3$

For $x = 3 + i$: $y = (3 + i) - 3 = i$ For $x = 3 - i$: $y = (3 - i) - 3 = -i$

Final Solution:

The solutions are: $(x, y) = (3 + i, i) \quad \text{and} \quad (x, y) = (3 - i, -i)$

Let me know if you want further clarification or a breakdown of any step! 😊

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Related Questions:

1. How is the quadratic formula derived?
2. What does the presence of imaginary roots indicate about a quadratic equation?
3. How can complex solutions be graphically represented?
4. What are some real-world applications of complex numbers?
5. Can we use substitution methods for other types of equations?

Tip: Always simplify equations before applying methods like the quadratic formula for easier calculations.