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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=12 2 x plus 2 y equals 12 −2x−2y=−22 negative 2 x minus 2 y equals negative 22 −x−y=−1 negative x minus y equals negative 1 2y=12

9 answers

To determine which equation creates a true equivalent system for System B, we need to analyze the equations given in the table.

From System B, we have the equation:

\[ x + y = 11 \]

To create an equivalent system, we can manipulate this equation.

  1. Multiply the entire equation by 2: \[ 2(x + y) = 2(11) \implies 2x + 2y = 22 \]

  2. Multiply the entire equation by -2: \[ -2(x + y) = -2(11) \implies -2x - 2y = -22 \]

  3. Rearranging the original equation: \[ x + y - 11 = 0 \implies -x - y = -11 \]

  4. Isolate \(y\): \[ y = 11 - x \] This does not yield a standard system equivalent.

Now, let's examine the response options given:

  1. 2x + 2y = 12 - This is not equivalent to \(x + y = 11\).
  2. −2x − 2y = −22 - This is equivalent to \(x + y = 11\) when divided by -2.
  3. −x − y = −1 - This is not equivalent to \(x + y = 11\).
  4. 2y = 12 - This simplifies to \(y = 6\) and does not represent the same relationship.

The correct answer is: −2x−2y=−22.

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 −14x−8y=8 negative 14 x minus 8 y equals 8 12x+12y=24 12 x plus 12 y equals 24 −12x−12y=24 negative 12 x minus 12 y equals 24 −14x−8y=−8

To determine which equation will make the systems equivalent, we start with the given equation from System A:

7x + 4y = -4

Next, we need to manipulate it into a new equivalent form. We can multiply the entire equation by some constant, typically a negative or positive integer, to create a system that maintains the same relationship.

Let's evaluate the answer options provided:

  1. −14x−8y=8

    • If we multiply the original equation \(7x + 4y = -4\) by 2, we get: \[ 2(7x + 4y) = 2(-4) \implies 14x + 8y = -8 \] So, this option does not match any equivalent form.

  2. 12x + 12y = 24

    • This does not relate to our original equation since multiplying by any constant would still keep it in the form of \(7x + 4y = -4\).

  3. −12x−12y=24

    • This again does not seem to follow from our equation.

  4. −14x−8y=−8

    • If we multiply the original equation \(7x + 4y = -4\) by -2, we get: \[ -2(7x + 4y) = -2(-4) \implies -14x - 8y = 8 \] which, when checked against this option, shows that we are not matching.

Notice the pattern that seems in focus, particularly with negative terms; flipping signs reverses the relationship unless the original term is correctly negated in equivalence.

Ultimately, the most promising option from our operations is:

−14x − 8y = −8

This matches our computed transformations based on multiplying by -2. Thus, the correct answer is:

−14x−8y=−8.

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 (−2,−4) left parenthesis negative 2 comma negative 4 right parenthesis (−16,2) left parenthesis negative 16 comma 2 right parenthesis

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)

First, we want to eliminate one of the variables by making the coefficients of \(x\) or \(y\) the same in both equations. Let's eliminate \(x\) by adjusting the second equation.

To do this, we can multiply Equation 2 by 3 so that the coefficients of \(x\) are the same:

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

This gives us:

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

Now we have the modified system:

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

Next, we will subtract Equation 1 from Equation 3 to eliminate \(x\):

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

This simplifies to:

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

So we have:

\[ 13y = -26 \]

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

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

Now that we have \(y\), we can substitute this value into Equation 2 to find \(x\):

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

This gives:

\[ x - 6 = -10 \]

Therefore:

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

The solution to the system of equations is:

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

So the correct response is:

\[ \text{(−4,−2)} \]

Use the elimination method to solve the linear-quadratic system y=x^2+10 , y=−7x−2 .(1 point) Responses (−4,26) left parenthesis negative 4 comma 26 right parenthesis (−3,19) left parenthesis negative 3 comma 19 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) and (−4,26)

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

  1. \(y = x^2 + 10\) (Equation 1) - a quadratic equation
  2. \(y = -7x - 2\) (Equation 2) - a linear equation

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

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

Now, let's rearrange this equation to form a standard quadratic equation by moving all terms to one side:

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

This simplifies to:

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

Next, we can factor the quadratic equation:

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

Thus, we find the solutions for \(x\):

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

Now, we can substitute these \(x\) values back into Equation 2 to find the corresponding \(y\) values.

For \(x = -3\):

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

So one solution is the point:

\((-3, 19)\).

For \(x = -4\):

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

So another solution is the point:

\((-4, 26)\).

The solutions to the linear-quadratic system are:

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

From the given responses, the correct answer is:

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

Use the elimination method to solve the linear-quadratic system y=x^2−2x−2 , y=2x−6 .(1 point) Responses (−2,−10) left parenthesis negative 2 comma negative 10 right parenthesis (2,−2) left parenthesis 2 comma negative 2 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) and (−2,2)

To solve the system of equations given by

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

we will set the two equations equal to each other since both are equal to \(y\):

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

Now, let's rearrange the equation to combine all terms on one side:

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

This simplifies to:

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

Next, we can factor this quadratic equation:

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

This gives us the solution:

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

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

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

So, the solution to the system is:

\[ (2, -2) \]

Now let's verify whether there are any other solutions. Since the equation \(x^2 - 4x + 4 = 0\) is a perfect square, it has only one distinct solution (a double root). Therefore, there are no other corresponding \(x\) values or points of intersection for the given system.

Thus, the correct response from the provided options is:

\((2, -2)\).