Question

Use the table to answer the 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?

(1 point)
Responses

2y=12
2 y equals 12

−2x−2y=−22
negative 2 x minus 2 y equals negative 22

2x+2y=12
2 x plus 2 y equals 12

−x−y=−1

Answer 1

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?


(1 point)
Responses

−12x−12y=24
negative 12 x minus 12 y equals 24

−14x−8y=−8
negative 14 x minus 8 equals negative 8

−14x−8y=8
negative 14 x minus 8 y equals 8

12x+12y=24

Answer 2

Question
Solve the following system of equations using the elimination method.

3x−4y=−4, x+3y=−10

(1 point)
Responses

(−4,−2)
left parenthesis negative 4 comma negative 2 right parenthesis

(−16,−2)
left parenthesis negative 16 comma negative 2 right parenthesis

(−16,2)
left parenthesis negative 16 comma 2 right parenthesis

(−2,−4)

Answer 3

Use the elimination method to solve the linear-quadratic system y=x2+10, y=−7x−2.(1 point)
Responses

(−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

(−4,26)
left parenthesis negative 4 comma 26 right parenthesis

(3,−23) and (4,−30)

Answer 4

Use the elimination method to solve the linear-quadratic system y=x2−2x−2, y=2x−6.(1 point)
Responses

(−2,−10)
left parenthesis negative 2 comma negative 10 right parenthesis

(−2,−2) and (−2,2)
left parenthesis negative 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis

(2,−2)
left parenthesis 2 comma negative 2 right parenthesis

(2,−2) and (−2,2)

Answer 5

To create a true equivalent system, we want an equation that does not change the solutions of the original equations provided in System A and System B.

System A has the equation:
- \( -x + y = 1 \)

System B has the equation:
- \( x + y = 11 \)

Let's analyze the options given one by one:

1. **\( 2y = 12 \)**: This simplifies to \( y = 6 \). This is not equivalent to the original equations.

2. **\( -2x - 2y = -22 \)**: This can be simplified to \( x + y = 11 \), which is the same as the equation in System B. This option works.

3. **\( 2x + 2y = 12 \)**: This simplifies to \( x + y = 6 \). This does not match the equations in the systems.

4. **\( -x - y = -1 \)**: This is equivalent to \( x + y = 1 \). This does not match the equations.

The only option that presents an equivalent equation is **\( -2x - 2y = -22 \)**, since it simplifies to one of the equations in System B \( x + y = 11 \).

Thus, the answer is:
**−2x−2y=−22** (negative 2 x minus 2 y equals negative 22)

Answer 6

To find the correct option that makes the systems equivalent, we need to determine what transformations can be applied to the equations without changing their solution sets.

Given:
- From System A: \( 7x + 4y = -4 \)
- From System B: \( 5x + 8y = 28 \)
- And one repeat of \( 7x + 4y = -4 \)

We need to find an additional equation in System B that represents the same relationship as the equations given in System A.

Let’s analyze each option one by one:

1. **Option: \( -12x - 12y = 24 \)**
Dividing every term by -12 gives us:
\[ x + y = -2 \]
This does not relate to either previous equation.

2. **Option: \( -14x - 8y = -8 \)**
Dividing every term by -2 gives us:
\[ 7x + 4y = 4 \]
This is not equivalent to the first equation and does not relate correctly.

3. **Option: \( -14x - 8y = 8 \)**
Dividing every term by -2 gives us:
\[ 7x + 4y = -4 \]
This is indeed equivalent to the first equation in System A.

4. **Option: \( 12x + 12y = 24 \)**
Dividing every term by 12 gives us:
\[ x + y = 2 \]
This does not relate to either previous equation.

The only option that produces an equation equivalent to the original equations in System A is:

**−14x−8y=8** (negative 14 x minus 8 y equals 8)

Answer 7

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

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

### Step 1: Align the equations
We want to eliminate one of the variables. Let's eliminate \( x \) by making the coefficients of \( x \) in both equations the same. We can multiply Equation 2 by \( 3 \):

\[
3(x + 3y) = 3(-10)
\]
This gives us:

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

### Step 2: Set up the elimination
Now we have the system:

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

### Step 3: Subtract the equations
Now we'll subtract Equation 1 from Equation 3 to eliminate \( x \):

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

This simplifies to:
\[
3x + 9y - 3x + 4y = -30 + 4
\]
\[
13y = -26
\]

### Step 4: Solve for \( y \)
Now, divide both sides by \( 13 \):

\[
y = -2
\]

### Step 5: Substitute \( y \) back to find \( x \)
Now, substitute \( y = -2 \) back into one of the original equations. Let's use Equation 2:

\[
x + 3(-2) = -10
\]
\[
x - 6 = -10
\]
\[
x = -10 + 6
\]
\[
x = -4
\]

### Conclusion
The solution to the system of equations is:

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

Thus, the answer is:

**(−4,−2)** (left parenthesis negative 4 comma negative 2 right parenthesis)

Answer 8

To solve the linear-quadratic system using substitution, we have two equations:

1. \( y = x^2 + 10 \) (Quadratic equation)
2. \( y = -7x - 2 \) (Linear equation)

### Step 1: Set the equations equal to each other

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

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

### Step 2: Rearrange the equation

Rearrange the equation to set it to zero:

\[
x^2 + 7x + 10 + 2 = 0
\]
\[
x^2 + 7x + 12 = 0
\]

### Step 3: Factor the quadratic equation

Next, we will factor the quadratic \( x^2 + 7x + 12 \):

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

### Step 4: Solve for \( x \)

Set each factor equal to zero:

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

### Step 5: Find the corresponding \( y \) values

Now we need to find the \( y \) values corresponding to each \( x \):

1. For \( x = -3 \):
\[
y = (-3)^2 + 10 = 9 + 10 = 19
\]
So one solution is \( (-3, 19) \).

2. For \( x = -4 \):
\[
y = (-4)^2 + 10 = 16 + 10 = 26
\]
So the second solution is \( (-4, 26) \).

### Conclusion

The solutions to the system of equations are:

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

Thus, the answer is:
**(−3,19) and (−4,26)** (left parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis)

Answer 9

To solve the linear-quadratic system using substitution or elimination, we have the following equations:

1. \( y = x^2 - 2x - 2 \) (Quadratic equation)
2. \( y = 2x - 6 \) (Linear equation)

### Step 1: Set the equations equal to each other

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

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

### Step 2: Rearrange the equation

Rearranging the equation to set it to zero gives us:

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

### Step 3: Factor the quadratic equation

The equation \( x^2 - 4x + 4 \) factors to:

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

### Step 4: Solve for \( x \)

Setting the factor equal to zero gives us:

\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

### Step 5: Find the corresponding \( y \) value

Now, substitute \( x = 2 \) back into either equation to find \( y \). We'll use the linear equation:

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

So, the solution to the system is:

\[
(2, -2)
\]

### Conclusion

The only solution to the system is:

**(2, −2)** (left parenthesis 2 comma negative 2 right parenthesis)