To solve these simultaneous equations:

$x^2 + y^2 = 17$ $2x + y = 6$

Let's go through it step-by-step.

Step 1: Express $y$ in terms of $x$ using the second equation

From $2x + y = 6$: $y = 6 - 2x$

Step 2: Substitute $y = 6 - 2x$ into the first equation

Substitute $y$ in $x^2 + y^2 = 17$: $x^2 + (6 - 2x)^2 = 17$

Expand $(6 - 2x)^2$: $x^2 + (36 - 24x + 4x^2) = 17$

Combine like terms: $5x^2 - 24x + 36 = 17$

Bring 17 to the left side: $5x^2 - 24x + 19 = 0$

Step 3: Solve the quadratic equation $5x^2 - 24x + 19 = 0$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = -24$, and $c = 19$.

Calculate the discriminant: $b^2 - 4ac = (-24)^2 - 4 \cdot 5 \cdot 19 = 576 - 380 = 196$

$x = \frac{24 \pm \sqrt{196}}{10} = \frac{24 \pm 14}{10}$

This gives: $x = \frac{24 + 14}{10} = 3.8 \quad \text{and} \quad x = \frac{24 - 14}{10} = 1.0$

Step 4: Find corresponding $y$ values for each $x$

1. For $x = 3.8$: $y = 6 - 2 \times 3.8 = 6 - 7.6 = -1.6$

2. For $x = 1.0$: $y = 6 - 2 \times 1.0 = 6 - 2 = 4$

Final Solution

The solutions to the simultaneous equations are: $(x, y) = (3.8, -1.6) \quad \text{and} \quad (x, y) = (1.0, 4)$

Would you like more details on any specific part of this solution?

Here are some related questions for further exploration:

1. How can we verify these solutions satisfy both equations?
2. What would happen if the quadratic equation had no real solutions?
3. Can we solve these equations using a graphical method?
4. How would the solution change if the second equation were $2x + y = k$ where $k \neq 6$?
5. What other methods could be used to solve this system?

Tip: In problems involving substitution in simultaneous equations, simplifying the equations as much as possible before substituting often makes the calculations easier.