Which of the following systems of equations has the solution (-2, 1)? This question requires you to show your work. (1 point) Responses 2x−y=−5 2 − = − 5 and x+2y=10 + 2 = 10 2 x minus y is equal to negative 5 and x plus 2 y is equal to 10 y=−2x−2 = − 2 − 2 and 2x−2y=5 2 − 2 = 5 y is equal to negative 2 x minus 2 and 2 x minus 2 y is equal to 5 x−4y=8 − 4 = 8 and 3x+10=−2y 3 + 10 = − 2 x minus 4 y is equal to 8 and 3 x plus 10 is equal to negative 2 y y=−32x−2 = − 3 2 − 2 and 2x−y=−5 2 − = − 5 y is equal to negative 3 over 2 x minus 2 and 2 x minus y is equal to negative 5
To find which system of equations has the solution (-2, 1), we substitute x = -2 and y = 1 into each system of equations and check if they hold true.
1. System: 2x - y = -5 and x + 2y = 10
- First equation: 2(-2) - 1 = -5
=> -4 - 1 = -5
=> -5 = -5 (holds true)
- Second equation: (-2) + 2(1) = 10
=> -2 + 2 = 10
=> 0 = 10 (does not hold true)
2. System: y = -2x - 2 and 2x - 2y = 5
- First equation: 1 = -2(-2) - 2
=> 1 = 4 - 2
=> 1 = 2 (does not hold true)
- Second equation: 2(-2) - 2(1) = 5
=> -4 - 2 = 5
=> -6 = 5 (does not hold true)
3. System: x - 4y = 8 and 3x + 10 = -2y
- First equation: (-2) - 4(1) = 8
=> -2 - 4 = 8
=> -6 = 8 (does not hold true)
- Second equation: 3(-2) + 10 = -2(1)
=> -6 + 10 = -2
=> 4 = -2 (does not hold true)
4. System: y = -3/2x - 2 and 2x - y = -5
- First equation: 1 = -3/2(-2) - 2
=> 1 = 3 + 2
=> 1 = 5 (does not hold true)
- Second equation: 2(-2) - 1 = -5
=> -4 - 1 = -5
=> -5 = -5 (holds true)
Therefore, the system of equations that has the solution (-2, 1) is:
2x - y = -5
y = -3/2x - 2