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.