We are tasked with solving the following system of simultaneous equations:

1. $y = 9 - 2x$
2. $y^2 + x^2 = 17$

Step 1: Substituting for $y$

Since we know that $y = 9 - 2x$, we can substitute this expression into the second equation:

$(9 - 2x)^2 + x^2 = 17$

Step 2: Expanding the equation

Expand $(9 - 2x)^2$:

$(9 - 2x)^2 = 81 - 36x + 4x^2$

Now, substitute this back into the equation:

$81 - 36x + 4x^2 + x^2 = 17$

Step 3: Simplify the equation

Combine like terms:

$5x^2 - 36x + 81 = 17$

Subtract 17 from both sides:

$5x^2 - 36x + 64 = 0$

Step 4: Solving the quadratic equation

We now have the quadratic equation:

$5x^2 - 36x + 64 = 0$

We'll solve this using the quadratic formula, which is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 5$, $b = -36$, and $c = 64$. Substituting these values into the formula:

$x = \frac{-(-36) \pm \sqrt{(-36)^2 - 4(5)(64)}}{2(5)}$

$x = \frac{36 \pm \sqrt{1296 - 1280}}{10}$

$x = \frac{36 \pm \sqrt{16}}{10}$

$x = \frac{36 \pm 4}{10}$

Step 5: Find the two solutions for $x$

Now, we calculate the two possible values for $x$:

1. $x = \frac{36 + 4}{10} = \frac{40}{10} = 4$
2. $x = \frac{36 - 4}{10} = \frac{32}{10} = \frac{16}{5}$

Step 6: Find the corresponding values of $y$

Using the equation $y = 9 - 2x$, we find the corresponding values of $y$:

1. If $x = 4$:

$y = 9 - 2(4) = 9 - 8 = 1$

2. If $x = \frac{16}{5}$:

$y = 9 - 2\left(\frac{16}{5}\right) = 9 - \frac{32}{5} = \frac{45}{5} - \frac{32}{5} = \frac{13}{5}$

Step 7: Conclusion

For $y > 2$, the solution is:

• $x = \frac{16}{5}$
• $y = \frac{13}{5}$

The other solution is:

• $x = 4$
• $y = 1$

Do you want more details or have any questions?

Here are five related questions:

1. How do you derive the quadratic formula?
2. What are other methods to solve simultaneous equations?
3. How can you tell which solution satisfies $y > 2$?
4. What are the applications of simultaneous equations in real life?
5. How can you check if a solution is valid for both equations?

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