Use the elimination method to solve the linear-quadratic system y=x^2−2x−2 , y=2x−6 .

Answer 1

To solve the linear-quadratic system given by the equations \( y = x^2 - 2x - 2 \) and \( y = 2x - 6 \) using the elimination method, we can start by setting the two expressions for \( y \) equal to each other:

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

Now, let's rearrange the equation to set everything to one side:

1. Move \( 2x \) and \( -6 \) from the right to the left:

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

2. Combine like terms:

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

3. Factor the quadratic equation:

\[ (x - 2)^2 = 0 \]

4. Solve for \( x \):

\[ x - 2 = 0 \implies x = 2 \]

Now that we have \( x = 2 \), we can substitute this value into one of the original equations to find \( y \). We can use the linear equation:

\[ y = 2x - 6 \]

Substituting \( x = 2 \):

\[ y = 2(2) - 6 \ y = 4 - 6 \ y = -2 \]

Thus, the solution to the system is:

\[ (x, y) = (2, -2) \]

To verify, we should check that this point also satisfies the quadratic equation \( y = x^2 - 2x - 2 \):

Substituting \( x = 2 \):

\[ y = 2^2 - 2(2) - 2 \ y = 4 - 4 - 2 \ y = -2 \]

Since both equations are satisfied, the solution to the system is:

\[ \boxed{(2, -2)} \]

Answer 2

Use the table to answer the question.

System A- 7x+4y=-4
5x+8y=28

System B- 7x+4y=-4
?Which of the following terms will make the systems equivalent?

Answer 3

To determine which term will make System B equivalent to System A, we start with the equations provided for both systems:

System A:

1. \( 7x + 4y = -4 \)
2. \( 5x + 8y = 28 \)

System B:

1. \( 7x + 4y = -4 \)
2. (to be determined)

For System B to be equivalent to System A, the second equation in System B must be a linear combination of the equations from System A. This means that the second equation in System B can be obtained by manipulating the second equation of System A.

Let’s analyze the second equation from System A: \( 5x + 8y = 28 \).

To find an equivalent equation, we can rewrite the equation in a format that can be compared or manipulated.

1. Multiply the entire equation by a constant factor or perform other linear combinations. However, the simplest way to find an equivalent equation is to keep the same set of solutions, which usually means scaling the equation by a constant.

If we were to multiply \( 5x + 8y = 28 \) by some constant, say \( k \), we would get:

\[ k(5x + 8y) = k(28) \ (5k)x + (8k)y = 28k \]

Thus, any equation that can be expressed in the form \( (5k)x + (8k)y = 28k \) (for any constant \( k \)) will be equivalent to \( 5x + 8y = 28 \).

Possible options for the second equation in System B:

• If you have specific options to choose from for the second equation to check, substitute them into the modified form \( 5kx + 8ky = 28k \) and see which one matches the form derived from System A.

In the absence of specific options presented, it's challenging to pinpoint an exact term. Generally, you will be looking for an equation with the same ratios for \( x \) and \( y \) coefficients (such as being scalar multiples) of \( 5 \) and \( 8 \) respectively, and the constant term being a multiple of \( 28 \).

Conclusion:

To make System B equivalent to System A, select a linear equation that maintains the same slope defined by the ratios of coefficients of \( x \) and \( y \) and the constant term as manipulated from System A's second equation.