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

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

2y=12
2 y equals 12

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

−x−y=−1

Answer 1

se 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

−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
negative 12 x minus 12 y equals 24

12x+12y=24
12 x plus 12 y equals 24

Answer 2

Solve the following system of equations using the elimination method.

3x−4y=−4, x+3y=−10

(1 point)
Responses

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

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

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

(−16,2)

Answer 3

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

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

(3,−23) and (4,−30)
left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis

(−3,19)
left parenthesis negative 3 comma 19 right parenthesis

(−3,19) and (−4,26)

Answer 4

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

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

(−2,−10)

Answer 5

To find an equivalent system to the given equations, we can start with the equations from System A and System B.

The equations in System A are:
1. \( -x + y = 1 \)

The equations in System B are:
1. \( x + y = 11 \)
2. \( x + y = 11 \) (repeated)

Now, if we look for equivalent forms of the first equation \( -x + y = 1 \):
- We can rearrange it to identify possible equivalent forms.
- Multiply through by -1 to get:
\[
x - y = -1
\]

For the second equation \( x + y = 11 \):
- Multiply through by 2 to obtain an equivalent equation:
\[
2x + 2y = 22
\]

However, the choices you provided were:

1. \( -2x - 2y = -22 \)
2. \( 2y = 12 \)
3. \( 2x + 2y = 12 \)
4. \( -x - y = -1 \)

To find an equivalent system, we can either manipulate these equations or find combinations that match.

Since we see that option 4, \( -x - y = -1 \), can be rearranged (by multiplying everything by -1) to get \( x + y = 1 \), it does not match our original system.

The option \( -2x - 2y = -22 \) simplifies to \( 2x + 2y = 22 \) when divided by -1, which is simply a scaled version of \( x + y = 11 \).

Therefore, the first option, \( -2x - 2y = -22 \), is indeed a valid equivalent statement.

So, the answer is:
**−2x−2y=−22**.

Answer 6

To determine which option would make the systems equivalent, we need to analyze the equations from both systems.

From your information, we have:

**System A:**
1. \(7x + 4y = -4\)

**System B:**
1. \(5x + 8y = 28\)
2. \(7x + 4y = -4\) (this is a repeat of the first equation)

To find a term that makes this system equivalent, we can look for a scalar multiple of the first equation, \(7x + 4y = -4\).

Here’s how we can manipulate this equation:

1. Multiply the entire equation \(7x + 4y = -4\) by a constant to create an equivalent equation.
2. Let’s say we multiply it by -2:
\[
-2(7x + 4y) = -2(-4)
\]
This simplifies to:
\[
-14x - 8y = 8
\]

Now we can look at the response options you provided:

1. \(-14x - 8y = -8\)
2. \(-14x - 8y = 8\)
3. \(-12x - 12y = 24\)
4. \(12x + 12y = 24\)

From our earlier calculation, we found that \( -14x - 8y = 8\) is an equivalent equation derived from \(7x + 4y = -4\).

Thus, the option that will make the systems equivalent is:

**−14x−8y=8**.

Answer 7

To solve the system of equations using the elimination method, we start with the two 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 by multiplying the equations, if necessary, such that when we add or subtract them, one variable cancels out.

**Step 2: Eliminate variable \(x\).**
Let's multiply Equation 2 by 3 so that the coefficients of \(x\) in both equations will be the same:

\[
3(x + 3y) = 3(-10)
\]
This gives us the modified Equation 2:
\[
3x + 9y = -30 \quad \text{(Equation 2 modified)}
\]

Now our system looks like this:

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

**Step 3: Subtract the two equations to eliminate \(x\).**
Now we can eliminate \(x\) by subtracting Equation 1 from the modified Equation 2:

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

This simplifies to:

\[
3x + 9y - 3x + 4y = -30 + 4
\]
\[
13y = -26
\]

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

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

**Step 5: Substitute back to find \(x\).**
Now that we have \(y\), we can substitute it back into either original equation to find \(x\). We'll use Equation 2:

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

Add \(6\) to both sides:

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

**Final Solution:**
The solution to the system of equations is:

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

Therefore, the correct response is:

**(−4,−2)**.

Answer 8

To solve the system of equations using the elimination method, we need to set up the equations we have:

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

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

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

**Step 1: Rearrange into standard form.**
To set this equation to 0, move all terms to one side:

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

**Step 2: Factor the quadratic equation.**

We now need to factor the quadratic equation \( x^2 + 7x + 12 \):

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

**Step 3: Solve for \( x \).**
Setting each factor to zero gives:

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

**Step 4: Solve for \( y \).**
Now we will substitute these \( x \) values back into either original equation to find the corresponding \( y \) values. We'll use Equation 2 for this.

- For \( x = -3 \):
\[
y = -7(-3) - 2 = 21 - 2 = 19
\]
So one point is \( (-3, 19) \).

- For \( x = -4 \):
\[
y = -7(-4) - 2 = 28 - 2 = 26
\]
So the other point is \( (-4, 26) \).

**Final Solution:**
The solutions to the system of equations are:

1. \( (-3, 19) \)
2. \( (-4, 26) \)

Therefore, the correct response is:

**(−3,19) and (−4,26)**.

Answer 9

To solve the linear-quadratic system using the elimination method, we look at the two equations:

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

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

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

**Step 1: Rearrange into standard form.**
Move all terms to one side of the equation:

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

This simplifies to:

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

**Step 2: Factor the quadratic equation.**

The equation can be factored as follows:

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

which can also be written as:

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

**Step 3: Solve for \( x \).**

Setting \( (x - 2)^2 = 0 \) gives us:

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

**Step 4: Solve for \( y \).**

Now we substitute \( x = 2 \) back into either original equation to find the corresponding \( y \):

Using Equation 2:

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

So we have one solution:

\[
(2, -2).
\]

Since both equations intersect at this point and there are no other solutions (because we have a repeated root), we conclude that the answers given do not yield multiple points.

**Final Solution:**
The only solution to the system of equations is:

\[
(2, -2).
\]

Therefore, the correct response is:

**(2,−2)**.