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We are given two simultaneous equations to solve:

\[
x^2 + y^2 = 41 \tag{1}
\]
\[
y = 2x - 3 \tag{2}
\]

### Step 1: Substitute Equation (2) into Equation (1)

From Equation (2), we know that \( y = 2x - 3 \). We substitute this expression for \( y \) into Equation (1):

\[
x^2 + (2x - 3)^2 = 41
\]

### Step 2: Expand and Simplify

Expand \( (2x - 3)^2 \):

\[
(2x - 3)^2 = (2x)^2 - 2(2x)(3) + (-3)^2 = 4x^2 - 12x + 9
\]

Substitute this back into the equation:

\[
x^2 + 4x^2 - 12x + 9 = 41
\]

Simplify:

\[
5x^2 - 12x + 9 = 41
\]

### Step 3: Solve the Quadratic Equation

Move all terms to one side:

\[
5x^2 - 12x + 9 - 41 = 0
\]

\[
5x^2 - 12x - 32 = 0
\]

Now solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(5)(-32)}}{2(5)}
\]

\[
x = \frac{12 \pm \sqrt{144 + 640}}{10}
\]

\[
x = \frac{12 \pm \sqrt{784}}{10}
\]

\[
x = \frac{12 \pm 28}{10}
\]

Thus, we have two solutions for \( x \):

\[
x = \frac{12 + 28}{10} = \frac{40}{10} = 4
\]

\[
x = \frac{12 - 28}{10} = \frac{-16}{10} = -1.6
\]

### Step 4: Find Corresponding \( y \) Values

For \( x = 4 \):

\[
y = 2(4) - 3 = 8 - 3 = 5
\]

For \( x = -1.6 \):

\[
y = 2(-1.6) - 3 = -3.2 - 3 = -6.2
\]

### Final Solutions

The solutions to the system are:

\[
(x, y) = (4, 5) \quad \text{or} \quad (x, y) = (-1.6, -6.2)
\]

Would you like more details on any of these steps, or do you have further questions?

Here are 5 related questions:
1. What other methods could be used to solve simultaneous equations, such as graphing?
2. How does the discriminant of a quadratic equation help in predicting the nature of the solutions?
3. Can simultaneous equations have more than two solutions? If so, under what circumstances?
4. What are some real-world applications of solving simultaneous equations like these?
5. How can you check whether a solution is correct when solving simultaneous equations?

**Tip:** Always substitute your solutions back into both original equations to verify that they satisfy both conditions.

Solve the simultaneous equations: x^2 + y^2 = 41 and y = 2x - 3.

This step-by-step solution shows how to solve the simultaneous equations x^2 + y^2 = 41 and y = 2x - 3 using substitution and the quadratic formula. Learn key algebraic concepts and theorems like quadratic equations and the quadratic formula. Suitable for high school students in Grades 9-12.

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