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=12 2 x plus 2 y equals 12 −x−y=−1 negative x minus y equals negative 1 2y=12 2 y equals 12 −2x−2y=−22 negative 2 x minus 2 y equals negative 22 Skip to navigation
Answers
Answer
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 12 x plus 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 negative 12 x minus 12 y equals 24 Skip to navigation
Answer
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) left parenthesis negative 16 comma 2 right parenthesis Skip to navigation page 15 of 15
Answer
Use the elimination method to solve the linear-quadratic system y=x2+10, y=−7x−2.(1 point) Responses (−3,19) and (−4,26) left parenthesis negative 3 comma 19 right parenthesis and 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 (−4,26)
Answer
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 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 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)
Answered by
GPT-4o mini
To create a true equivalent system, you can manipulate the original equations while maintaining their overall equality.
Let's analyze the equations provided:
1. For System A: \( -x + y = 1 \)
2. For System B: \( x + y = 11 \)
To check the options given:
1. **\(2x + 2y = 12\)**: This is equivalent to both equations when combined, since if we multiply the second equation \(x + y = 11\) by 2, we get \(2x + 2y = 22\), not 12.
2. **\(-x - y = -1\)**: If we add both original equations, we get 0 = 12, which does not help us. However, if we take the first equation and modify it correctly, it leads to \(-x - y = -1\), which is equivalent when simplified.
3. **\(2y = 12\)**: This is equivalent to \(y = 6\), not directly equivalent to either original equation.
4. **\(-2x - 2y = -22\)**: If we multiply the second equation by -2, we correctly get \(-2x - 2y = -22\), making this a true equivalent as well.
So, two possible options are correct:
- **\(-x - y = -1\)**
- **\(-2x - 2y = -22\)**
However, of those provided, the most directly related equation from the original system while maintaining the structure of the original equations is \(-x - y = -1\).
Thus, the choice that correctly supports the formation of an equivalent system is:
**−x−y=−1**.
Let's analyze the equations provided:
1. For System A: \( -x + y = 1 \)
2. For System B: \( x + y = 11 \)
To check the options given:
1. **\(2x + 2y = 12\)**: This is equivalent to both equations when combined, since if we multiply the second equation \(x + y = 11\) by 2, we get \(2x + 2y = 22\), not 12.
2. **\(-x - y = -1\)**: If we add both original equations, we get 0 = 12, which does not help us. However, if we take the first equation and modify it correctly, it leads to \(-x - y = -1\), which is equivalent when simplified.
3. **\(2y = 12\)**: This is equivalent to \(y = 6\), not directly equivalent to either original equation.
4. **\(-2x - 2y = -22\)**: If we multiply the second equation by -2, we correctly get \(-2x - 2y = -22\), making this a true equivalent as well.
So, two possible options are correct:
- **\(-x - y = -1\)**
- **\(-2x - 2y = -22\)**
However, of those provided, the most directly related equation from the original system while maintaining the structure of the original equations is \(-x - y = -1\).
Thus, the choice that correctly supports the formation of an equivalent system is:
**−x−y=−1**.
Answered by
GPT-4o mini
To determine which of the given options can make the system equivalent, we need to analyze System A and System B:
1. For System A: \( 7x + 4y = -4 \)
2. For System B: \( 5x + 8y = 28 \)
To find the equivalent system, we need to manipulate either of the equations without changing their equality.
### Check the Options:
1. **\( 12x + 12y = 24 \)**: This does not relate directly to either equation from the systems provided.
2. **\(-14x - 8y = -8\)**: To check if this is equivalent, let's rearrange it:
- Multiplying the second equation from System B \((5x + 8y = 28)\) by -1 gives us \(-5x - 8y = -28\). This is not equivalent to the equation here.
3. **\(-14x - 8y = 8\)**: From our previous transformation, we can see this does not match either system's equations directly for equivalency.
4. **\(-12x - 12y = 24\)**: Dividing this equation by -12 gives \(x + y = -2\), which does not relate directly to either equation.
### Exploring Equivalent Transformations:
One approach is scaling the original equations.
If we multiply \(7x + 4y = -4\) from System A by -2:
\[
-2(7x + 4y = -4) \Rightarrow -14x - 8y = 8
\]
This matches **\(-14x - 8y = 8\)**.
### Conclusion
The equation that will create an equivalent system is:
**−14x−8y=8**.
1. For System A: \( 7x + 4y = -4 \)
2. For System B: \( 5x + 8y = 28 \)
To find the equivalent system, we need to manipulate either of the equations without changing their equality.
### Check the Options:
1. **\( 12x + 12y = 24 \)**: This does not relate directly to either equation from the systems provided.
2. **\(-14x - 8y = -8\)**: To check if this is equivalent, let's rearrange it:
- Multiplying the second equation from System B \((5x + 8y = 28)\) by -1 gives us \(-5x - 8y = -28\). This is not equivalent to the equation here.
3. **\(-14x - 8y = 8\)**: From our previous transformation, we can see this does not match either system's equations directly for equivalency.
4. **\(-12x - 12y = 24\)**: Dividing this equation by -12 gives \(x + y = -2\), which does not relate directly to either equation.
### Exploring Equivalent Transformations:
One approach is scaling the original equations.
If we multiply \(7x + 4y = -4\) from System A by -2:
\[
-2(7x + 4y = -4) \Rightarrow -14x - 8y = 8
\]
This matches **\(-14x - 8y = 8\)**.
### Conclusion
The equation that will create an equivalent system is:
**−14x−8y=8**.
Answered by
GPT-4o mini
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 for elimination
We will eliminate one variable by making the coefficients of \(x\) or \(y\) the same in both equations. Let's eliminate \(x\).
To do this, we can multiply Equation 2 by 3 (to match the coefficient of \(x\) in Equation 1):
\[
3(x + 3y) = 3(-10)
\]
This gives us:
\[
3x + 9y = -30 \quad \text{(Equation 3)}
\]
### Step 2: Rewrite the system
Now we have:
1. \( 3x - 4y = -4 \) (Equation 1)
2. \( 3x + 9y = -30 \) (Equation 3)
### Step 3: Eliminate \(x\)
Now we can subtract Equation 1 from Equation 3:
\[
(3x + 9y) - (3x - 4y) = -30 - (-4)
\]
This simplifies to:
\[
3x + 9y - 3x + 4y = -30 + 4
\]
or
\[
13y = -26
\]
### Step 4: Solve for \(y\)
Now, divide both sides by 13:
\[
y = -2
\]
### Step 5: Substitute \(y\) back to find \(x\)
Now we substitute \(y = -2\) back into one of the original equations. We can use Equation 2:
\[
x + 3(-2) = -10
\]
This simplifies to:
\[
x - 6 = -10
\]
Add 6 to both sides:
\[
x = -4
\]
### Final Solution
The solution to the system of equations is \( (x, y) = (-4, -2) \).
Thus, the answer is:
**(−4,−2)**.
1. \( 3x - 4y = -4 \) (Equation 1)
2. \( x + 3y = -10 \) (Equation 2)
### Step 1: Align the equations for elimination
We will eliminate one variable by making the coefficients of \(x\) or \(y\) the same in both equations. Let's eliminate \(x\).
To do this, we can multiply Equation 2 by 3 (to match the coefficient of \(x\) in Equation 1):
\[
3(x + 3y) = 3(-10)
\]
This gives us:
\[
3x + 9y = -30 \quad \text{(Equation 3)}
\]
### Step 2: Rewrite the system
Now we have:
1. \( 3x - 4y = -4 \) (Equation 1)
2. \( 3x + 9y = -30 \) (Equation 3)
### Step 3: Eliminate \(x\)
Now we can subtract Equation 1 from Equation 3:
\[
(3x + 9y) - (3x - 4y) = -30 - (-4)
\]
This simplifies to:
\[
3x + 9y - 3x + 4y = -30 + 4
\]
or
\[
13y = -26
\]
### Step 4: Solve for \(y\)
Now, divide both sides by 13:
\[
y = -2
\]
### Step 5: Substitute \(y\) back to find \(x\)
Now we substitute \(y = -2\) back into one of the original equations. We can use Equation 2:
\[
x + 3(-2) = -10
\]
This simplifies to:
\[
x - 6 = -10
\]
Add 6 to both sides:
\[
x = -4
\]
### Final Solution
The solution to the system of equations is \( (x, y) = (-4, -2) \).
Thus, the answer is:
**(−4,−2)**.
Answered by
GPT-4o mini
To solve the system of equations given by the linear equation and the quadratic equation using the elimination method, we have:
1. \(y = x^2 + 10\) (Equation 1)
2. \(y = -7x - 2\) (Equation 2)
### Step 1: Set the equations equal to each other
Since both equations equal \(y\), we can set them equal to each other:
\[
x^2 + 10 = -7x - 2
\]
### Step 2: Rearrange the equation to form a standard quadratic equation
To solve for \(x\), let's rearrange this equation:
\[
x^2 + 7x + 10 + 2 = 0
\]
This simplifies to:
\[
x^2 + 7x + 12 = 0
\]
### Step 3: Factor the quadratic equation
Next, we can factor the quadratic equation:
\[
(x + 3)(x + 4) = 0
\]
Setting each factor to zero gives us:
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]
### Step 4: Find the corresponding \(y\) values
Now that we have the \(x\) values, we can plug them back into either of the original equations to find the corresponding \(y\) values. We'll use Equation 2 for convenience.
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 other solution is \((-4, 26)\).
### Final Solutions
The solutions to the system of equations are:
1. \((-3, 19)\)
2. \((-4, 26)\)
So the correct response is:
**(−3, 19) and (−4, 26)**.
1. \(y = x^2 + 10\) (Equation 1)
2. \(y = -7x - 2\) (Equation 2)
### Step 1: Set the equations equal to each other
Since both equations equal \(y\), we can set them equal to each other:
\[
x^2 + 10 = -7x - 2
\]
### Step 2: Rearrange the equation to form a standard quadratic equation
To solve for \(x\), let's rearrange this equation:
\[
x^2 + 7x + 10 + 2 = 0
\]
This simplifies to:
\[
x^2 + 7x + 12 = 0
\]
### Step 3: Factor the quadratic equation
Next, we can factor the quadratic equation:
\[
(x + 3)(x + 4) = 0
\]
Setting each factor to zero gives us:
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]
### Step 4: Find the corresponding \(y\) values
Now that we have the \(x\) values, we can plug them back into either of the original equations to find the corresponding \(y\) values. We'll use Equation 2 for convenience.
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 other solution is \((-4, 26)\).
### Final Solutions
The solutions to the system of equations are:
1. \((-3, 19)\)
2. \((-4, 26)\)
So the correct response is:
**(−3, 19) and (−4, 26)**.
Answered by
GPT-4o mini
To solve the 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
Because both equations are equal to \( y \), we can set them equal to each other:
\[
x^2 - 2x - 2 = 2x - 6
\]
### Step 2: Rearrange to form a standard quadratic equation
Now, let’s rearrange the equation:
\[
x^2 - 2x - 2 - 2x + 6 = 0
\]
Combine like terms:
\[
x^2 - 4x + 4 = 0
\]
### Step 3: Factor the quadratic equation
Next, we can factor this quadratic equation:
\[
(x - 2)(x - 2) = 0
\]
This can also be written as:
\[
(x - 2)^2 = 0
\]
Setting the factored form to zero gives us:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
### Step 4: Find the corresponding \(y\) value
Now that we have \(x = 2\), we can substitute this back into either of the original equations to find \(y\). Let’s use Equation 2:
\[
y = 2(2) - 6
\]
Calculating gives:
\[
y = 4 - 6 = -2
\]
### Final Solution
The solution to the system of equations is:
\[
(2, -2)
\]
Since the quadratic equation has a repeated root, this will be the only solution to the system.
Thus, the correct response is:
**(2,−2)**.
1. \( y = x^2 - 2x - 2 \) (Equation 1)
2. \( y = 2x - 6 \) (Equation 2)
### Step 1: Set the equations equal to each other
Because both equations are equal to \( y \), we can set them equal to each other:
\[
x^2 - 2x - 2 = 2x - 6
\]
### Step 2: Rearrange to form a standard quadratic equation
Now, let’s rearrange the equation:
\[
x^2 - 2x - 2 - 2x + 6 = 0
\]
Combine like terms:
\[
x^2 - 4x + 4 = 0
\]
### Step 3: Factor the quadratic equation
Next, we can factor this quadratic equation:
\[
(x - 2)(x - 2) = 0
\]
This can also be written as:
\[
(x - 2)^2 = 0
\]
Setting the factored form to zero gives us:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
### Step 4: Find the corresponding \(y\) value
Now that we have \(x = 2\), we can substitute this back into either of the original equations to find \(y\). Let’s use Equation 2:
\[
y = 2(2) - 6
\]
Calculating gives:
\[
y = 4 - 6 = -2
\]
### Final Solution
The solution to the system of equations is:
\[
(2, -2)
\]
Since the quadratic equation has a repeated root, this will be the only solution to the system.
Thus, the correct response is:
**(2,−2)**.
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