Solving Simultaneous Equations Using Elimination Practice

Question

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Question
Use the elimination method to solve the linear-quadratic system y=x2+2x−4
, y=−x−6
.(1 point)
The solutions are (−1,−5)
and (
).
Remaining Attempts : 3
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Answer 1

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

1. \( y = x^2 + 2x - 4 \)
2. \( y = -x - 6 \)

Since both equations are equal to \( y \), we can set them equal to each other:

\[ x^2 + 2x - 4 = -x - 6 \]

Now, let's move all terms to one side of the equation to set it to zero:

\[ x^2 + 2x + x - 4 + 6 = 0 \]

This simplifies to:

\[ x^2 + 3x + 2 = 0 \]

Next, we can factor this quadratic equation:

\[ (x + 1)(x + 2) = 0 \]

Setting each factor to zero gives us the solutions for \( x \):

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

Now we can substitute these \( x \) values back into either of the original equations to find the corresponding \( y \) values. We can use the second equation since it's simpler:

1. For \( x = -1 \): \[ y = -(-1) - 6 = 1 - 6 = -5 \] Therefore, one solution is \( (-1, -5) \).

2. For \( x = -2 \): \[ y = -(-2) - 6 = 2 - 6 = -4 \] Therefore, the second solution is \( (-2, -4) \).

Now, we have two solutions: \( (-1, -5) \) and \( (-2, -4) \).

To complete the problem, input the missing solution:

The solutions are (-1, -5) and (-2, -4).