Question

Use the elimination method to solve the linear-quadratic system y=x2+10 , 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 (−4,26) left parenthesis negative 4 comma 26 right parenthesis (−3,19) left parenthesis negative 3 comma 19 right parenthesis (−3,19) and (−4,26) left parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis

Answer 1

To solve the system of equations using the elimination method, we first need to set the equations equal to each other by substituting for \( y \).

1. The equations given are: \[ y = x^2 + 10 \] \[ y = -7x - 2 \]

2. Set the right sides equal to each other: \[ x^2 + 10 = -7x - 2 \]

3. Rearranging the equation gives: \[ x^2 + 7x + 12 = 0 \]

4. Next, we can factor the quadratic equation: \[ (x + 3)(x + 4) = 0 \]

5. Setting each factor to zero gives: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

6. Now we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We can use \( y = -7x - 2 \).

For \( x = -3 \): \[ y = -7(-3) - 2 = 21 - 2 = 19 \] So one solution is \((-3, 19)\).

For \( x = -4 \): \[ y = -7(-4) - 2 = 28 - 2 = 26 \] So the other solution is \((-4, 26)\).

Therefore, the solutions to the system of equations are: \[ (-3, 19) \quad \text{and} \quad (-4, 26) \]

So the correct response is: \[ \text{ } (-3, 19) \text{ and } (-4, 26) \]