2-Variable Equations & Inequalities Online Practice

Question

Complete this assessment to review what you’ve learned. It will not count toward your grade.
15 of 1815 of 18 Items

Question
Use the elimination method to solve the linear-quadratic system y=x2+x−42
, y=−x+6
.(1 point)
Responses

(8,−2)
and (−6,12)
left parenthesis 8 comma negative 2 right parenthesis and left parenthesis negative 6 comma 12 right parenthesis

(6,0)
left parenthesis 6 comma 0 right parenthesis

(−8,14)
left parenthesis negative 8 comma 14 right parenthesis

(−8,14)
and (6,0)
left parenthesis negative 8 comma 14 right parenthesis and left parenthesis 6 comma 0 right parenthesis
Remaining Attempts : 3
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Answer 1

To solve the system of equations using the elimination method, we have:

1. \( y = x^2 + x - 42 \) (quadratic equation)
2. \( y = -x + 6 \) (linear equation)

First, we can set the two equations equal to each other since they both equal \( y \):

\[ x^2 + x - 42 = -x + 6 \]

Now, let's rearrange this equation:

\[ x^2 + x + x - 42 - 6 = 0 \] \[ x^2 + 2x - 48 = 0 \]

Next, we will factor the quadratic equation \( x^2 + 2x - 48 = 0 \).

To factor, we need two numbers that multiply to \(-48\) and add up to \(2\). These numbers are \(8\) and \(-6\). So we can rewrite the equation as:

\[ (x + 8)(x - 6) = 0 \]

Setting each factor to zero gives us:

1. \( x + 8 = 0 \) \( \Rightarrow x = -8 \)
2. \( x - 6 = 0 \) \( \Rightarrow x = 6 \)

Now that we have \( x \) values, we can find the corresponding \( y \) values using the second equation \( y = -x + 6 \).

For \( x = -8 \): \[ y = -(-8) + 6 = 8 + 6 = 14 \] So, one solution is \( (-8, 14) \).

For \( x = 6 \): \[ y = -(6) + 6 = -6 + 6 = 0 \] Thus, the second solution is \( (6, 0) \).

The solutions to the system of equations are:

• \( (-8, 14) \)
• \( (6, 0) \)

Now, looking at your responses:

• \( (8, -2) \) and \( (-6, 12) \)
• \( (6, 0) \)
• \( (-8, 14) \)
• \( (-8, 14) \) and \( (6, 0) \)

The correct answer is:

\( (-8, 14) \) and \( (6, 0) \).