Use the table to answer the question.

Question

System A System B
−x+y=1
x+y=11 x+y=11
?

Which of the following shows the equation that creates a true equivalent system?

Responses

−x−y=−1

−2x−2y=−22

2x+2y=12

2y=12

Answer 1

To determine which option creates a true equivalent system with the original equations, we need to check which option can be derived from the given equations.

We have:

System A:

1. \( -x + y = 1 \)
2. \( x + y = 11 \)

First, let's manipulate the second equation in System A for consideration:

From Equation 2: \[ x + y = 11 \] We can multiply this entire equation by 2 to find an equivalent equation: \[ 2(x + y) = 2(11) \implies 2x + 2y = 22 \quad (1) \]

Now let's analyze each option provided:

1. −x−y=−1

   • This is equivalent to \( x + y = 1 \), which does not relate to our equations.

2. −2x−2y=−22

   • Dividing the equation by -2 gives us \( x + y = 11 \). This matches one of our original equations.

3. 2x+2y=12

   • This simplifies to \( x + y = 6 \), which does not match our equations.

4. 2y=12

   • Dividing by 2 gives \( y = 6 \). This does not create an equivalent system since it does not match either equation.

The correct answer that transforms into an equivalent system is: −2x−2y=−22

This option \( (-2x - 2y = -22) \) correctly corresponds to the equation \( (x + y = 11) \).

Answer 2

Use the table to answer the question.

System A System B
7x+4y=−4

5x+8y=28
7x+4y=−4

?

Which of the following terms will make the systems equivalent?

Responses:

−12x−12y=24

−14x−8y=8

12x+12y=24

−14x−8y=−8

Answer 3

To determine which equation will create an equivalent system with the given equations in System A, we need to look at the existing equations:

System A:

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

For an equivalent system, we can manipulate one of the existing equations to derive a new equation.

Step 1: Manipulate the first equation

We can multiply the first equation by a constant to find an equivalent equation. Let's multiply \( 7x + 4y = -4 \) by -2:

\[ -2(7x + 4y) = -2(-4) \] This gives us: \[ -14x - 8y = 8 \quad \text{ (1)} \]

Step 2: Check the options

Now let's examine each option to see if it matches the result:

1. −12x−12y=24

   • This does not match our manipulated equation.

2. −14x−8y=8

   • This matches the derived equation (1).

3. 12x+12y=24

   • This does not match our manipulated equation.

4. −14x−8y=−8

   • This is not equivalent to our derived equation as it has a different constant on the right side.

Conclusion

The equation that will create an equivalent system is: −14x−8y=8

This option correctly corresponds to one of the equations in System A after manipulation.

Answer 4

Solve the following system of equations using the elimination method.

3x−4y=−4, x+3y=−10

Responses

(-2,-4)

(−16,2)

(−16,−2)

(−4,−2)

Answer 5

To solve the system of equations using the elimination method, we have the following two equations:

1. \( 3x - 4y = -4 \) (Equation 1)
2. \( x + 3y = -10 \) (Equation 2)

Step 1: Align the equations

We can either multiply one or both equations to eliminate one variable. Let's try to eliminate \(x\) by making the coefficients of \(x\) in both equations the same.

From Equation 2, let's multiply it by 3 to match the coefficient of \(x\) in Equation 1:

\[ 3(x + 3y) = 3(-10) \]

This gives us:

\[ 3x + 9y = -30 \quad (Equation 3) \]

Step 2: Subtract the equations

Now we have:

1. \( 3x - 4y = -4 \) (Equation 1)
2. \( 3x + 9y = -30 \) (Equation 3)

Next, we subtract Equation 1 from Equation 3:

\[ (3x + 9y) - (3x - 4y) = -30 - (-4) \]

This simplifies to:

\[ 3x + 9y - 3x + 4y = -30 + 4 \]

Combine the like terms:

\[ 13y = -26 \]

Step 3: Solve for \(y\)

Now, we can solve for \(y\):

\[ y = \frac{-26}{13} = -2 \]

Step 4: Substitute \(y\) back to find \(x\)

Now that we have \(y\), we can substitute this value back into either of the original equations to find \(x\). We'll use Equation 2:

\[ x + 3(-2) = -10 \]

This simplifies to:

\[ x - 6 = -10 \]

Now, solve for \(x\):

\[ x = -10 + 6 = -4 \]

Step 5: Solution

Therefore, the solution to the system of equations is:

\[ (x, y) = (-4, -2) \]

Response

The correct response is: (−4,−2)

Answer 6

Use the elimination method to solve the linear-quadratic system y=x^2+10 , y=−7x−2 .(1 point)
Responses

(3,−23) and (4,−30)

(−3,19)

(−4,26)

(−3,19) and (−4,26)

Answer 7

To solve the linear-quadratic system of equations using the elimination method, we have the following two equations:

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

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

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

Step 1: Rearranging the equation

Now, let's rearrange the equation to set it to 0:

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

Step 2: Factoring the quadratic equation

Next, we can factor the quadratic:

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

Step 3: Finding the values of \(x\)

Setting each factor equal to zero gives us:

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

Step 4: Finding the corresponding \(y\) values

Now we need to find the corresponding \(y\) values for each \(x\). We can use either original equation; we will use Equation 2:

1. For \( x = -3 \):

\[ y = -7(-3) - 2 = 21 - 2 = 19 \]

So one solution is \( (-3, 19) \).

2. For \( x = -4 \):

\[ y = -7(-4) - 2 = 28 - 2 = 26 \]

So the second solution is \( (-4, 26) \).

Step 5: Solutions

Thus, the solutions to the system are:

\[ (-3, 19) \quad \text{and} \quad (-4, 26) \]

Response

The correct response is: (−3,19) and (−4,26)

Answer 8

Use the elimination method to solve the linear-quadratic system y=x^2−2x−2 , y=2x−6 .(1 point)
Responses

(−2,−2) and (−2,2)

(−2,−10)

(2,−2)

(2,−2) and (−2,2)

Answer 9

To solve the linear-quadratic system of equations using the elimination method, we have:

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

Step 1: Set the equations equal to each other

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

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

Step 2: Rearranging the equation

Now, let's rearrange the equation to set it to 0:

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

Step 3: Factoring the quadratic equation

Now we can factor the quadratic:

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

Step 4: Finding the values of \(x\)

Setting the factor equal to zero gives us:

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

Step 5: Finding the corresponding \(y\) value

Now, we need to find the corresponding \(y\) value for \(x = 2\). We can use either of the original equations; we'll use Equation 2:

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

So, the only solution to the system is \( (2, -2) \).

Step 6: Conclusion

Thus, the solution to the linear-quadratic system is:

\[ (2, -2) \]

Response

The correct response is: (2,−2).