We are given the following system of simultaneous equations:

1. $x + y = 9$
2. $x^2 - 3xy + 2y^2 = 0$

Step 1: Express $x$ in terms of $y$

From the first equation, solve for $x$:

$x = 9 - y$

Step 2: Substitute $x = 9 - y$ into the second equation

Now, substitute $x = 9 - y$ into the second equation:

$(9 - y)^2 - 3(9 - y)y + 2y^2 = 0$

Expand each term: $(9 - y)^2 = 81 - 18y + y^2$ $-3(9 - y)y = -27y + 3y^2$

Thus, the equation becomes: $(81 - 18y + y^2) + (-27y + 3y^2) + 2y^2 = 0$

Simplify: $81 - 18y + y^2 - 27y + 3y^2 + 2y^2 = 0$ $81 - 45y + 6y^2 = 0$

Step 3: Solve the quadratic equation

We have the quadratic equation: $6y^2 - 45y + 81 = 0$

Divide the entire equation by 3 to simplify: $2y^2 - 15y + 27 = 0$

Now, solve this quadratic equation using the quadratic formula: $y = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(27)}}{2(2)}$ $y = \frac{15 \pm \sqrt{225 - 216}}{4}$ $y = \frac{15 \pm \sqrt{9}}{4}$ $y = \frac{15 \pm 3}{4}$

So, the two solutions for $y$ are: $y = \frac{15 + 3}{4} = \frac{18}{4} = 4.5$ $y = \frac{15 - 3}{4} = \frac{12}{4} = 3$

Step 4: Find the corresponding $x$ values

For $y = 4.5$, substitute into $x = 9 - y$: $x = 9 - 4.5 = 4.5$

For $y = 3$, substitute into $x = 9 - y$: $x = 9 - 3 = 6$

Final Solutions:

The solutions to the system are: $(x, y) = (4.5, 4.5) \quad \text{and} \quad (x, y) = (6, 3)$

Would you like more details on any step or have any questions?

Here are 5 related questions to explore further:

1. How can you verify the solutions to a system of simultaneous equations?
2. What are the general methods for solving nonlinear simultaneous equations?
3. Can a system of quadratic equations have more than two solutions?
4. How does the discriminant of a quadratic equation affect the number of solutions?
5. What are the different forms of simultaneous equations, and how do you approach solving each type?

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