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Question

Use the substitution method to solve this linear-quadratic system. y=x^2-4x-41 & y=4

Answer 1

To solve the system of equations using the substitution method, we start with the two equations given:

1. \( y = x^2 - 4x - 41 \)
2. \( y = 4 \)

Since the second equation tells us that \( y \) is equal to 4, we can substitute \( 4 \) for \( y \) in the first equation. This will allow us to solve for \( x \).

Substituting \( y = 4 \) into the first equation gives:

\[ 4 = x^2 - 4x - 41 \]

Now, we'll rearrange this equation to set it to zero:

\[ x^2 - 4x - 41 - 4 = 0 \]

This simplifies to:

\[ x^2 - 4x - 45 = 0 \]

Now we need to factor or use the quadratic formula to solve for \( x \). We will first try to factor:

We need two numbers that multiply to \(-45\) and add to \(-4\). The numbers \(-9\) and \(5\) satisfy this.

So we can factor the equation as:

\[ (x - 9)(x + 5) = 0 \]

Setting each factor to zero gives us:

1. \( x - 9 = 0 \) → \( x = 9 \)
2. \( x + 5 = 0 \) → \( x = -5 \)

Now we have two \( x \)-values: \( x = 9 \) and \( x = -5 \).

Next, we'll find the corresponding \( y \)-values using the second equation \( y = 4 \).

So no matter what \( x \) is, \( y = 4 \).

Therefore, the solution pairs to the system of equations are:

\[ (9, 4) \quad \text{and} \quad (-5, 4) \]

The final solutions are:

\[ (9, 4) \quad \text{and} \quad (-5, 4) \]

Answer 2

Way shorter please

Answer 3

Sure! Here's the solution in a more concise format:

1. Given the equations:

   • \( y = x^2 - 4x - 41 \)
   • \( y = 4 \)

2. Substitute \( y = 4 \) into the first equation: \[ 4 = x^2 - 4x - 41 \]

3. Rearrange to find \( x \): \[ x^2 - 4x - 45 = 0 \]

4. Factor the quadratic: \[ (x - 9)(x + 5) = 0 \]

5. Solve for \( x \): \[ x = 9 \quad \text{or} \quad x = -5 \]

6. Corresponding \( y \): \[ y = 4 \]

7. Solutions are: \[ (9, 4) \quad \text{and} \quad (-5, 4) \]

Answer 4

Shorter please

Answer 5

Sure! Here’s a very brief version:

1. Given:

   • \( y = x^2 - 4x - 41 \)
   • \( y = 4 \)

2. Substitute \( y = 4 \): \[ 4 = x^2 - 4x - 41 \] \[ x^2 - 4x - 45 = 0 \]

3. Factor: \[ (x - 9)(x + 5) = 0 \implies x = 9, -5 \]

4. So, \( y = 4 \) for both.

5. Solutions: \[ (9, 4) \quad \text{and} \quad (-5, 4) \]

Answer 6

shorter

Answer 7

Absolutely! Here’s the shortest version:

1. Solve \( 4 = x^2 - 4x - 41 \): \[ x^2 - 4x - 45 = 0 \quad \Rightarrow \quad (x - 9)(x + 5) = 0 \]

2. Solutions: \( x = 9, -5 \) with \( y = 4 \).

3. Final answers: \[ (9, 4), (-5, 4) \]