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
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
y=−2x−2
=
−
2
−
2
and 2x−2y=5
To find which system of equations has the solution (-2, 1), we need to substitute the values of x = -2 and y = 1 into each system and see which one satisfies both equations.
1) For 2x - y = -5 and x + 2y = 10:
2(-2) - 1 = -5
-4 - 1 = -5
-5 = -5 (satisfied)
(-2) + 2(1) = 10
-2 + 2 = 10
0 = 10 (not satisfied)
2) For x - 4y = 8 and 3x + 10 = -2y:
(-2) - 4(1) = 8
-2 - 4 = 8
-6 = 8 (not satisfied)
3(-2) + 10 = -2(1)
-6 + 10 = -2
4 = -2 (not satisfied)
3) For y = -3/2x - 2 and 2x - y = -5:
1 = -3/2(-2) - 2
1 = 3 + 2
1 = 5 (not satisfied)
2(-2) - 1 = -5
-4 - 1 = -5
-5 = -5 (satisfied)
4) For y = -2x - 2 and 2x - 2y = 5:
1 = -2(-2) - 2
1 = 4 - 2
1 = 2 (not satisfied)
2(-2) - 2(1) = 5
-4 - 2 = 5
-6 = 5 (not satisfied)
Therefore, the system of equations that has the solution (-2, 1) is y = -3/2x - 2 and 2x - y = -5.