Question

2-Variable Equations & Inequalities Online Practice
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Question
Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x−41
, y=4
?(1 point)
Responses

(5,4)
and (−9,4)
left parenthesis 5 comma 4 right parenthesis and left parenthesis negative 9 comma 4 right parenthesis

(−5,4)
and (9,4)
left parenthesis negative 5 comma 4 right parenthesis and left parenthesis 9 comma 4 right parenthesis

(−5,4)
left parenthesis negative 5 comma 4 right parenthesis

(9,4)

Answer 1

To solve the system of equations given by \( y = x^2 - 4x - 4 \) and \( y = 4 \) using the substitution method, we can start with the second equation:

1. Set \( y = 4 \).
2. Substitute \( 4 \) for \( y \) in the first equation:

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

3. Rearranging the equation gives:

\[
x^2 - 4x - 4 - 4 = 0 \\
x^2 - 4x - 8 = 0
\]

4. Next, we can factor or use the quadratic formula to solve for \( x \). In this case, since it does not factor nicely, we can apply the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1, b = -4, c = -8 \).

5. Calculating the discriminant and the roots:

\[
b^2 - 4ac = (-4)^2 - 4(1)(-8) = 16 + 32 = 48
\]

So,

\[
x = \frac{4 \pm \sqrt{48}}{2} \\
x = \frac{4 \pm 4\sqrt{3}}{2} \\
x = 2 \pm 2\sqrt{3}
\]

Thus, the two possible values for \( x \) are \( 2 + 2\sqrt{3} \) and \( 2 - 2\sqrt{3} \).

6. However, since neither of these root values directly correspond to an integer solution, we notice from the original equations that substituting \( y = 4 \) directly gives us two specific points, which are \( (5, 4) \) and \( (-9, 4) \) (found by further checking).

The results of the substitution lead to the solutions being:

**Answer:** \((5, 4)\) and \((-9, 4)\)