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Use the substitution method to solve this linear�quadratic system. 𝑦 = 𝑥

2 − 4𝑥 − 41 & 𝑦 =

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To solve the linear-quadratic system using the substitution method, we have the following two equations:

  1. \( y = x^2 - 4x - 41 \) (quadratic equation)
  2. \( y = mx + b \) (linear equation, but we need it to proceed)

Since the second equation is not fully provided in your question, let’s assume it represents a linear equation of the format \( y = mx + b \). If you can provide the specific values for \( m \) and \( b \), we can solve it more accurately. However, I will demonstrate the process generally.

Steps to solve the system:

  1. Starting with the first equation \( y = x^2 - 4x - 41 \), we can substitute this expression for \( y \) into the linear equation.

  2. Therefore, substituting gives us: \[ mx + b = x^2 - 4x - 41 \]

  3. Rearranging this equation to set it to zero results in: \[ x^2 - (4 + m)x - (41 + b) = 0 \]

  4. This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \). Now let's denote:

    • \( a = 1 \)
    • \( b = -(4 + m) \)
    • \( c = -(41 + b) \)

  5. To find the solutions for \( x \), use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

  6. Substituting the known values: \[ x = \frac{(4 + m) \pm \sqrt{(4 + m)^2 - 4(1)(-(41 + b))}}{2 \cdot 1} \]

  7. Once \( x \) is found, substitute \( x \) back into the original equation for \( y \) (or you can use the linear equation).

  8. This will yield your \( y \) values accordingly.

Example Calculation:

Let's say you have a specific linear equation \( y = 2x + 1 \).

Then, substituting yields: \[ 2x + 1 = x^2 - 4x - 41 \] Rearranging: \[ x^2 - 6x - 42 = 0 \]

Solving this using the quadratic formula: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-42)}}{2(1)} = \frac{6 \pm \sqrt{36 + 168}}{2} = \frac{6 \pm \sqrt{204}}{2} = \frac{6 \pm 2\sqrt{51}}{2} = 3 \pm \sqrt{51} \]

Find corresponding y:

Substituting back to find \( y \): \[ y = 2(3 \pm \sqrt{51}) + 1 = 6 \pm 2\sqrt{51} + 1 = 7 \pm 2\sqrt{51} \]

You will have two sets of solutions:

  1. \( (3 + \sqrt{51}, 7 + 2\sqrt{51}) \)
  2. \( (3 - \sqrt{51}, 7 - 2\sqrt{51}) \)

Please provide the linear equation if you would like a more specific solution!