Which of the following uses the substitution method to solve linear-quadratic system y=x^2−4x−41, y=4

Question

a) -5,4 and 9,4
b) 5,4 and -9,4
c) -5,4
d) 9,4

Answer 1

To solve the linear-quadratic system \( y = x^2 - 4x - 41 \) and \( y = 4 \) using the substitution method, we can set the two equations equal to each other:

1. Start with the equations: \[ y = x^2 - 4x - 41 \] \[ y = 4 \]

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

3. Rearrange this into standard form: \[ x^2 - 4x - 41 - 4 = 0 \] \[ x^2 - 4x - 45 = 0 \]

4. Now we can factor the quadratic equation: \[ (x - 9)(x + 5) = 0 \]

5. Setting each factor equal to zero gives us the possible \( x \)-values: \[ x - 9 = 0 \quad \Rightarrow \quad x = 9 \] \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]

6. Now, we substitute these \( x \)-values back into \( y = 4 \) to get the corresponding \( y \)-values:

   • For \( x = 9 \), \( y = 4 \) gives the point \( (9, 4) \).
   • For \( x = -5 \), \( y = 4 \) gives the point \( (-5, 4) \).

Therefore, the system has two solutions: \( (9, 4) \) and \( (-5, 4) \).

Thus, the correct answer is: a) -5,4 and 9,4.