To solve the system of simultaneous equations:

$x^2 + y^2 = 36$ $x = 2y + 6$

we'll proceed as follows:

Step 1: Substitute $x = 2y + 6$ into the first equation.

Substitute $x = 2y + 6$ into $x^2 + y^2 = 36$:

$(2y + 6)^2 + y^2 = 36$

Step 2: Expand and simplify the equation.

Expand $(2y + 6)^2$:

$(2y + 6)^2 = 4y^2 + 24y + 36$

So, the equation becomes:

$4y^2 + 24y + 36 + y^2 = 36$

Combine like terms:

$5y^2 + 24y + 36 = 36$

Step 3: Move all terms to one side of the equation.

Subtract 36 from both sides:

$5y^2 + 24y = 0$

Step 4: Factor the equation.

Factor out $y$ from the equation:

$y(5y + 24) = 0$

Step 5: Solve for $y$.

The equation $y(5y + 24) = 0$ gives two solutions:

1. $y = 0$
2. $5y + 24 = 0$

For the second equation:

$5y = -24 \quad \Rightarrow \quad y = -\frac{24}{5} = -4.8$

So, the solutions for $y$ are $y = 0$ and $y = -4.8$.

Step 6: Find the corresponding $x$ values.

Using $x = 2y + 6$:

• If $y = 0$:
  $x = 2(0) + 6 = 6$

• If $y = -4.8$:
  $x = 2(-4.8) + 6 = -9.6 + 6 = -3.6$

Final Solutions:

The solutions to the system are:

1. $(x, y) = (6, 0)$
2. $(x, y) = (-3.6, -4.8)$

Would you like more details or have any questions?

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Here are 8 related questions you might find useful:

1. What is the geometric interpretation of the solutions to this system of equations?
2. How do you solve simultaneous linear equations using substitution?
3. What are some alternative methods to solve simultaneous equations?
4. How does the solution set change if the first equation is $x^2 + y^2 = r^2$ instead of 36?
5. What is the role of factoring in solving quadratic equations?
6. How can we check if our solutions are correct?
7. What happens if the equations have no real solution?
8. How would you graph the two equations to find the points of intersection?

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