Question

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