Question
A system of equations is given below.
5x-2y=3
3x-y=4
Which of these procedures will eliminate a variable in one of the equations in the system above?
Responses
Multiply the first equation by 2 then add the result to the second equation.
Multiply the first equation by 2 then add the result to the second equation.
Multiply the first equation by –2 then add the result to the second equation.
Multiply the first equation by –2 then add the result to the second equation.
Multiply the second equation by 2 then add the result to the first equation.
Multiply the second equation by 2 then add the result to the first equation.
Multiply the second equation by –2 then add the result to the first equation.
Multiply the second equation by –2 then add the result to the first equation.
Which coordinate point does not belong in its solution set?
Responses
(0, 0)
(0, 0)
(2, 2)
(2, 2)
(1, 4)
(1, 4)
(4, 1)
5x-2y=3
3x-y=4
Which of these procedures will eliminate a variable in one of the equations in the system above?
Responses
Multiply the first equation by 2 then add the result to the second equation.
Multiply the first equation by 2 then add the result to the second equation.
Multiply the first equation by –2 then add the result to the second equation.
Multiply the first equation by –2 then add the result to the second equation.
Multiply the second equation by 2 then add the result to the first equation.
Multiply the second equation by 2 then add the result to the first equation.
Multiply the second equation by –2 then add the result to the first equation.
Multiply the second equation by –2 then add the result to the first equation.
Which coordinate point does not belong in its solution set?
Responses
(0, 0)
(0, 0)
(2, 2)
(2, 2)
(1, 4)
(1, 4)
(4, 1)
Answers
GPT-4o mini
To eliminate a variable in the given system of equations, we can manipulate the equations to make the coefficients of one of the variables opposites so that they cancel each other out when added. The system of equations is:
1. \( 5x - 2y = 3 \)
2. \( 3x - y = 4 \)
Let's analyze the provided responses for eliminating a variable:
1. **Multiply the first equation by 2 then add the result to the second equation.**
- This will not effectively eliminate a variable as the resulting equations won't have coefficients that can cancel each other out.
2. **Multiply the first equation by –2 then add the result to the second equation.**
- This equation would look like \( -10x + 4y = -6 \) added to \( 3x - y = 4 \). This could eliminate a variable, specifically \( y \).
3. **Multiply the second equation by 2 then add the result to the first equation.**
- This equation would look like \( 6x - 2y = 8 \) added to \( 5x - 2y = 3 \). This may also not help in eliminating a variable, as both have the same \( y \) coefficients but won't lead to the cancellation.
4. **Multiply the second equation by -2 then add the result to the first equation.**
- This yields \( -6x + 2y = -8 \), and adding that to \( 5x - 2y = 3 \) would effectively eliminate \( y \).
Thus, the best option for eliminating a variable is to **multiply the second equation by –2 then add the result to the first equation.**
Now for the second part, we need to find out which coordinate does not belong in the solution set. We will substitute each coordinate into the original equations to see if they satisfy both equations.
1. **(0, 0)**:
- \( 5(0) - 2(0) = 0 \) (does not satisfy \( = 3 \))
- \( 3(0) - 0 = 0 \) (does not satisfy \( = 4 \))
2. **(2, 2)**:
- \( 5(2) - 2(2) = 10 - 4 = 6 \) (does not satisfy \( = 3 \))
- \( 3(2) - 2 = 6 - 2 = 4 \) (satisfies)
3. **(1, 4)**:
- \( 5(1) - 2(4) = 5 - 8 = -3 \) (does not satisfy \( = 3 \))
- \( 3(1) - 4 = 3 - 4 = -1 \) (does not satisfy \( = 4 \))
4. **(4, 1)**:
- \( 5(4) - 2(1) = 20 - 2 = 18 \) (does not satisfy \( = 3 \))
- \( 3(4) - 1 = 12 - 1 = 11 \) (does not satisfy \( = 4 \))
Since none of the points satisfy both equations, we note that (0, 0) does not satisfy either of the equations.
The coordinate point that does not belong in the solution set is **(0, 0)**.
1. \( 5x - 2y = 3 \)
2. \( 3x - y = 4 \)
Let's analyze the provided responses for eliminating a variable:
1. **Multiply the first equation by 2 then add the result to the second equation.**
- This will not effectively eliminate a variable as the resulting equations won't have coefficients that can cancel each other out.
2. **Multiply the first equation by –2 then add the result to the second equation.**
- This equation would look like \( -10x + 4y = -6 \) added to \( 3x - y = 4 \). This could eliminate a variable, specifically \( y \).
3. **Multiply the second equation by 2 then add the result to the first equation.**
- This equation would look like \( 6x - 2y = 8 \) added to \( 5x - 2y = 3 \). This may also not help in eliminating a variable, as both have the same \( y \) coefficients but won't lead to the cancellation.
4. **Multiply the second equation by -2 then add the result to the first equation.**
- This yields \( -6x + 2y = -8 \), and adding that to \( 5x - 2y = 3 \) would effectively eliminate \( y \).
Thus, the best option for eliminating a variable is to **multiply the second equation by –2 then add the result to the first equation.**
Now for the second part, we need to find out which coordinate does not belong in the solution set. We will substitute each coordinate into the original equations to see if they satisfy both equations.
1. **(0, 0)**:
- \( 5(0) - 2(0) = 0 \) (does not satisfy \( = 3 \))
- \( 3(0) - 0 = 0 \) (does not satisfy \( = 4 \))
2. **(2, 2)**:
- \( 5(2) - 2(2) = 10 - 4 = 6 \) (does not satisfy \( = 3 \))
- \( 3(2) - 2 = 6 - 2 = 4 \) (satisfies)
3. **(1, 4)**:
- \( 5(1) - 2(4) = 5 - 8 = -3 \) (does not satisfy \( = 3 \))
- \( 3(1) - 4 = 3 - 4 = -1 \) (does not satisfy \( = 4 \))
4. **(4, 1)**:
- \( 5(4) - 2(1) = 20 - 2 = 18 \) (does not satisfy \( = 3 \))
- \( 3(4) - 1 = 12 - 1 = 11 \) (does not satisfy \( = 4 \))
Since none of the points satisfy both equations, we note that (0, 0) does not satisfy either of the equations.
The coordinate point that does not belong in the solution set is **(0, 0)**.