Solving Simultaneous Equations Using Elimination Quick Check

Question

4 of 54 of 5 Items

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)

Answer 1

To solve the system of equations \(y = x^2 + 10\) and \(y = -7x - 2\) using the elimination method, we first set the two equations equal to each other, since both are equal to \(y\):

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

Now, rearranging this equation:

2. \(x^2 + 7x + 12 = 0\)

Next, we can factor the quadratic equation:

3. \((x + 3)(x + 4) = 0\)

So we have two solutions for \(x\):

4. \(x = -3\) and \(x = -4\)

Now, we will substitute these values back into one of the original equations to find the corresponding \(y\) values. We'll 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 \((-4, 26)\).

Thus, the solutions to the system are \((-3, 19)\) and \((-4, 26)\).

The correct response from the choices you provided is:

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