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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

9 answers

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.

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

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.

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)

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).

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)

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).

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)

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).