For which system is (-2,-3) a solution?
a) 4x+2y=7, x-y=-5
b) 3x=2y, x+y=5
c) x-y=1, 3x-2y=0
d) x-1=2y, 2x-y=-7
I think it is C.
substitute (-2,-3) into:
c) x-y=1, 3x-2y=0
x-y = -2-(-3) = -2 +3 =1
3x-2y = 3(-2) - 2(-3) = -6 - (-6) = 0
So it is C.
To determine which system of equations (-2, -3) is a solution for, we can substitute the values of x and y into each equation and see if the equations hold true.
a) For the first system, substitute x = -2 and y = -3 into each equation:
4(-2) + 2(-3) = 7
-8 - 6 = 7
-14 ≠ 7
(-2) - (-3) = -5
-2 + 3 = -5
1 ≠ -5
Therefore, (-2, -3) is not a solution for the first system.
b) For the second system, substitute x = -2 and y = -3 into each equation:
3(-2) = 2(-3)
-6 = -6
(-2) + (-3) = 5
-5 = 5
Therefore, (-2, -3) is not a solution for the second system.
c) For the third system, substitute x = -2 and y = -3 into each equation:
(-2) - (-3) = 1
-2 + 3 = 1
3(-2) - 2(-3) = 0
-6 + 6 = 0
Therefore, (-2, -3) is a solution for the third system.
d) For the fourth system, substitute x = -2 and y = -3 into each equation:
(-2) - 1 = 2(-3)
-3 = -7
2(-2) - (-3) = -7
-4 + 3 = -7
Therefore, (-2, -3) is not a solution for the fourth system.
Thus, the correct answer is c) x - y = 1, 3x - 2y = 0, as (-2, -3) satisfies both equations.