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What operation needs to be performed to go from the equation 2x+7y=14 to the equivalent equation 4x+14y=28 .(1 point) The equat...Question
What operation needs to be performed to go from the equation 2x+7y=14
to the equivalent equation 4x+14y=28
.(1 point)
The equation 2x+7y=14
should be multiplied by
to produce the equivalent equation 4x+14y=28
.
to the equivalent equation 4x+14y=28
.(1 point)
The equation 2x+7y=14
should be multiplied by
to produce the equivalent equation 4x+14y=28
.
Answers
Answer
Solve the following system of equations using the elimination method.
2x−3y=14
, 4x+3y=10
(1 point)
2x−3y=14
, 4x+3y=10
(1 point)
Answer
Use the substitution method to solve the system 2x+3y=4 , −4x−6y=−8 .(1 point)
Responses
(−1,2)
left parenthesis negative 1 comma 2 right parenthesis
no solution
no solution
infinite solutions
infinite solutions
(2−1)
Use the substitution method to solve the system x=y−4 , x+8y=2 .(1 point)
Responses
no solution
no solution
(2/3,−3 1/3)
left parenthesis Start Fraction 2 over 3 End Fraction comma negative 3 Start Fraction 1 over 3 End Fraction right parenthesis
infinite solutions
infinite solutions
(−3 1/3,2/3)
Which of the following shows a system of equations equivalent to 2x+7y=14 and 4x+5y=10 ?(1 point)
Responses
2x+7y=14 and 2x+5y=5
2 x plus 7 y equals 14 and 2 x plus 5 y equals 5
4x+14y=14 and 4x+5y=10
4 x plus 14 y equals 14 and 4 x plus 5 y equals 10
2x+7y=14 and 8x+10y=20
2 x plus 7 y equals 14 and 8 x plus 10 y equals 20
4x+14y=28 and 8x+10y=10
Why is the system of equations 2x+7y=14 and 4x+5y=10 equivalent to the system of equations 4x+14y=28 and 8x+10y=20 ?(1 point)
Responses
All equations share the point (0,2).
All equations share the point left parenthesis 0 comma 2 right parenthesis .
Both equations, 2x+7y=14 and 4x+5y=10, can be multiplied by 12 to produce the equivalent system of equations.
Both equations, 2 x plus 7 y equals 14 and 4 x plus 5 y equals 10 , can be multiplied by Start Fraction 1 over 2 End Fraction to produce the equivalent system of equations.
The intercepts of the equations 2x+7y=14 and 4x+14y=28 are (0,2) and (7,0).
The intercepts of the equations 2 x plus 7 y equals 14 and 4 x plus 14 y equals 28 are left parenthesis 0 comma 2 right parenthesis and left parenthesis 7 comma 0 right parenthesis .
The intercepts of both 4x+5y=10 and 8x+10y=20 are (0,2.5) and (2,0).
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 2 right parenthesis
(−16,−2)
left parenthesis negative 16 comma negative 2 right parenthesis
(−2,−4)
Responses
(−1,2)
left parenthesis negative 1 comma 2 right parenthesis
no solution
no solution
infinite solutions
infinite solutions
(2−1)
Use the substitution method to solve the system x=y−4 , x+8y=2 .(1 point)
Responses
no solution
no solution
(2/3,−3 1/3)
left parenthesis Start Fraction 2 over 3 End Fraction comma negative 3 Start Fraction 1 over 3 End Fraction right parenthesis
infinite solutions
infinite solutions
(−3 1/3,2/3)
Which of the following shows a system of equations equivalent to 2x+7y=14 and 4x+5y=10 ?(1 point)
Responses
2x+7y=14 and 2x+5y=5
2 x plus 7 y equals 14 and 2 x plus 5 y equals 5
4x+14y=14 and 4x+5y=10
4 x plus 14 y equals 14 and 4 x plus 5 y equals 10
2x+7y=14 and 8x+10y=20
2 x plus 7 y equals 14 and 8 x plus 10 y equals 20
4x+14y=28 and 8x+10y=10
Why is the system of equations 2x+7y=14 and 4x+5y=10 equivalent to the system of equations 4x+14y=28 and 8x+10y=20 ?(1 point)
Responses
All equations share the point (0,2).
All equations share the point left parenthesis 0 comma 2 right parenthesis .
Both equations, 2x+7y=14 and 4x+5y=10, can be multiplied by 12 to produce the equivalent system of equations.
Both equations, 2 x plus 7 y equals 14 and 4 x plus 5 y equals 10 , can be multiplied by Start Fraction 1 over 2 End Fraction to produce the equivalent system of equations.
The intercepts of the equations 2x+7y=14 and 4x+14y=28 are (0,2) and (7,0).
The intercepts of the equations 2 x plus 7 y equals 14 and 4 x plus 14 y equals 28 are left parenthesis 0 comma 2 right parenthesis and left parenthesis 7 comma 0 right parenthesis .
The intercepts of both 4x+5y=10 and 8x+10y=20 are (0,2.5) and (2,0).
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 2 right parenthesis
(−16,−2)
left parenthesis negative 16 comma negative 2 right parenthesis
(−2,−4)
Answered by
GPT-4o mini
To go from the equation \( 2x + 7y = 14 \) to the equivalent equation \( 4x + 14y = 28 \), you need to multiply the entire equation by \( 2 \).
So, the statement would be:
The equation \( 2x + 7y = 14 \) should be multiplied by \( 2 \) to produce the equivalent equation \( 4x + 14y = 28 \).
So, the statement would be:
The equation \( 2x + 7y = 14 \) should be multiplied by \( 2 \) to produce the equivalent equation \( 4x + 14y = 28 \).
Answered by
GPT-4o mini
To solve the system of equations using the elimination method, we have the following two equations:
1. \( 2x - 3y = 14 \) \quad (Equation 1)
2. \( 4x + 3y = 10 \) \quad (Equation 2)
First, let's manipulate these equations to eliminate one of the variables. Since the coefficients of \( y \) in the two equations have opposite signs, we can add the two equations directly after multiplying Equation 1 by \( 1 \) (to keep it as it is) and Equation 2 by \( 1 \) (to keep it as it is).
### Step 1: Prepare to eliminate \( y \)
We can add the two equations together:
\[
(2x - 3y) + (4x + 3y) = 14 + 10
\]
### Step 2: Combine the equations
This results in:
\[
2x + 4x - 3y + 3y = 14 + 10
\]
\[
6x = 24
\]
### Step 3: Solve for \( x \)
Now, divide both sides by \( 6 \):
\[
x = \frac{24}{6} = 4
\]
### Step 4: Substitute \( x \) back to find \( y \)
Now that we have \( x = 4 \), substitute this value back into either of the original equations. We'll use Equation 1:
\[
2(4) - 3y = 14
\]
This simplifies to:
\[
8 - 3y = 14
\]
### Step 5: Solve for \( y \)
Subtract \( 8 \) from both sides:
\[
-3y = 14 - 8
\]
\[
-3y = 6
\]
Now, divide both sides by \( -3 \):
\[
y = \frac{6}{-3} = -2
\]
### Final Solution
The solution to the system of equations is:
\[
(x, y) = (4, -2)
\]
1. \( 2x - 3y = 14 \) \quad (Equation 1)
2. \( 4x + 3y = 10 \) \quad (Equation 2)
First, let's manipulate these equations to eliminate one of the variables. Since the coefficients of \( y \) in the two equations have opposite signs, we can add the two equations directly after multiplying Equation 1 by \( 1 \) (to keep it as it is) and Equation 2 by \( 1 \) (to keep it as it is).
### Step 1: Prepare to eliminate \( y \)
We can add the two equations together:
\[
(2x - 3y) + (4x + 3y) = 14 + 10
\]
### Step 2: Combine the equations
This results in:
\[
2x + 4x - 3y + 3y = 14 + 10
\]
\[
6x = 24
\]
### Step 3: Solve for \( x \)
Now, divide both sides by \( 6 \):
\[
x = \frac{24}{6} = 4
\]
### Step 4: Substitute \( x \) back to find \( y \)
Now that we have \( x = 4 \), substitute this value back into either of the original equations. We'll use Equation 1:
\[
2(4) - 3y = 14
\]
This simplifies to:
\[
8 - 3y = 14
\]
### Step 5: Solve for \( y \)
Subtract \( 8 \) from both sides:
\[
-3y = 14 - 8
\]
\[
-3y = 6
\]
Now, divide both sides by \( -3 \):
\[
y = \frac{6}{-3} = -2
\]
### Final Solution
The solution to the system of equations is:
\[
(x, y) = (4, -2)
\]
Answered by
GPT-4o mini
### 1. Substitution Method for the First System of Equations
**Given System:**
1. \( 2x + 3y = 4 \)
2. \( -4x - 6y = -8 \)
Let's simplify the second equation:
\[
-4x - 6y = -8 \implies 4x + 6y = 8 \quad (\text{just multiply by } -1)
\]
We can see that the second equation is just double the first equation:
\[
4x + 6y = 8 \text{ can be obtained by multiplying } (2x + 3y = 4) \text{ by } 2.
\]
This means the two equations represent the same line, hence there are **infinite solutions**.
- **Response:** **infinite solutions**
---
### 2. Substitution Method for the Second System of Equations
**Given System:**
1. \( x = y - 4 \)
2. \( x + 8y = 2 \)
Substituting \( x \) from the first equation into the second:
\[
(y - 4) + 8y = 2
\]
Combine like terms:
\[
y - 4 + 8y = 2 \implies 9y - 4 = 2
\]
Add \( 4 \) to both sides:
\[
9y = 6 \implies y = \frac{6}{9} = \frac{2}{3}
\]
Now substitute back to find \( x \):
\[
x = y - 4 = \frac{2}{3} - 4 = \frac{2}{3} - \frac{12}{3} = \frac{-10}{3}
\]
So the answer to the second system is:
- **Response:** **(2/3, −3 1/3)**
---
### 3. Equivalent System of Equations
**Given Equations:**
- \( 2x + 7y = 14 \)
- \( 4x + 5y = 10 \)
To check for equivalence, let's identify each option:
1. **\( 2x + 7y = 14 \) and \( 2x + 5y = 5 \)** - Not equivalent.
2. **\( 4x + 14y = 14 \) and \( 4x + 5y = 10 \)** - Not equivalent.
3. **\( 2x + 7y = 14 \) and \( 8x + 10y = 20 \)** - Not equivalent (not properly scaled).
4. **\( 4x + 14y = 28 \) and \( 8x + 10y = 20 \)** - Correct, because you could arrive at these by multiplying.
So the correct response would be:
- **Response:** \( 4x + 14y = 28 \) and \( 8x + 10y = 20 \)
---
### 4. Why is the system equivalent?
The first system is equivalent to the second system because they represent the same relationships. The simplest option is:
- Both equations, \( 2x + 7y = 14 \) and \( 4x + 5y = 10 \), can be multiplied by \( 2 \) to produce another equivalent form.
- **Response:** Both equations, \( 2x + 7y = 14 \) and \( 4x + 5y = 10 \), can be multiplied by \( \frac{1}{2} \) to produce the equivalent system of equations.
---
### 5. Elimination Method for the Last System of Equations
**Given System:**
1. \( 3x - 4y = -4 \)
2. \( x + 3y = -10 \)
Let's eliminate \( x \) by manipulating the second equation:
First, rearranging the second equation:
\[
x = -10 - 3y
\]
Now substitute \( x \) in the first equation:
\[
3(-10 - 3y) - 4y = -4
\]
Distribute \( 3 \):
\[
-30 - 9y - 4y = -4
\]
Combine like terms:
\[
-30 - 13y = -4
\]
Add \( 30 \) to both sides:
\[
-13y = 26
\]
So:
\[
y = -2
\]
Now substitute back to find \( x \):
\[
x = -10 - 3(-2) = -10 + 6 = -4
\]
So the solution to the system is:
- **Response:** **(−4, −2)**
---
Please let me know if you need any further assistance!
**Given System:**
1. \( 2x + 3y = 4 \)
2. \( -4x - 6y = -8 \)
Let's simplify the second equation:
\[
-4x - 6y = -8 \implies 4x + 6y = 8 \quad (\text{just multiply by } -1)
\]
We can see that the second equation is just double the first equation:
\[
4x + 6y = 8 \text{ can be obtained by multiplying } (2x + 3y = 4) \text{ by } 2.
\]
This means the two equations represent the same line, hence there are **infinite solutions**.
- **Response:** **infinite solutions**
---
### 2. Substitution Method for the Second System of Equations
**Given System:**
1. \( x = y - 4 \)
2. \( x + 8y = 2 \)
Substituting \( x \) from the first equation into the second:
\[
(y - 4) + 8y = 2
\]
Combine like terms:
\[
y - 4 + 8y = 2 \implies 9y - 4 = 2
\]
Add \( 4 \) to both sides:
\[
9y = 6 \implies y = \frac{6}{9} = \frac{2}{3}
\]
Now substitute back to find \( x \):
\[
x = y - 4 = \frac{2}{3} - 4 = \frac{2}{3} - \frac{12}{3} = \frac{-10}{3}
\]
So the answer to the second system is:
- **Response:** **(2/3, −3 1/3)**
---
### 3. Equivalent System of Equations
**Given Equations:**
- \( 2x + 7y = 14 \)
- \( 4x + 5y = 10 \)
To check for equivalence, let's identify each option:
1. **\( 2x + 7y = 14 \) and \( 2x + 5y = 5 \)** - Not equivalent.
2. **\( 4x + 14y = 14 \) and \( 4x + 5y = 10 \)** - Not equivalent.
3. **\( 2x + 7y = 14 \) and \( 8x + 10y = 20 \)** - Not equivalent (not properly scaled).
4. **\( 4x + 14y = 28 \) and \( 8x + 10y = 20 \)** - Correct, because you could arrive at these by multiplying.
So the correct response would be:
- **Response:** \( 4x + 14y = 28 \) and \( 8x + 10y = 20 \)
---
### 4. Why is the system equivalent?
The first system is equivalent to the second system because they represent the same relationships. The simplest option is:
- Both equations, \( 2x + 7y = 14 \) and \( 4x + 5y = 10 \), can be multiplied by \( 2 \) to produce another equivalent form.
- **Response:** Both equations, \( 2x + 7y = 14 \) and \( 4x + 5y = 10 \), can be multiplied by \( \frac{1}{2} \) to produce the equivalent system of equations.
---
### 5. Elimination Method for the Last System of Equations
**Given System:**
1. \( 3x - 4y = -4 \)
2. \( x + 3y = -10 \)
Let's eliminate \( x \) by manipulating the second equation:
First, rearranging the second equation:
\[
x = -10 - 3y
\]
Now substitute \( x \) in the first equation:
\[
3(-10 - 3y) - 4y = -4
\]
Distribute \( 3 \):
\[
-30 - 9y - 4y = -4
\]
Combine like terms:
\[
-30 - 13y = -4
\]
Add \( 30 \) to both sides:
\[
-13y = 26
\]
So:
\[
y = -2
\]
Now substitute back to find \( x \):
\[
x = -10 - 3(-2) = -10 + 6 = -4
\]
So the solution to the system is:
- **Response:** **(−4, −2)**
---
Please let me know if you need any further assistance!
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