Use the elimination method to solve the linear-quadratic system y=x2+10

Question

, y=−7x−2
.(1 point)
Responses

(3,−23)
and (4,−30)
left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis

(−3,19)
and (−4,26)
left parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis

(−3,19)
left parenthesis negative 3 comma 19 right parenthesis

(−4,26)
left parenthesis negative 4 comma 26 right parenthesis

Answer 1

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

1. \( y = x^2 + 10 \)
2. \( y = -7x - 2 \)

We can set these two equations equal to each other since they both equal \( y \):

\[ x^2 + 10 = -7x - 2 \]

Next, we will rearrange this equation to bring all terms to one side:

\[ x^2 + 7x + 10 + 2 = 0 \]

\[ x^2 + 7x + 12 = 0 \]

Now we will factor the quadratic equation:

\[ (x + 3)(x + 4) = 0 \]

Set each factor equal to zero:

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x + 4 = 0 \) → \( x = -4 \)

Now we will find the corresponding \( y \) values for each \( x \) value by substituting back into either of the original equations. We will use the second equation \( y = -7x - 2 \).

1. For \( x = -3 \): \[ y = -7(-3) - 2 = 21 - 2 = 19 \] This gives the point (-3, 19).

2. For \( x = -4 \): \[ y = -7(-4) - 2 = 28 - 2 = 26 \] This gives the point (-4, 26).

Thus, the solutions to the system are:

• \((-3, 19)\)
• \((-4, 26)\)

So the answer is:

\((-3, 19)\) and \((-4, 26)\).