One method is to set each equation in the form y=mx+b, set those two equations equal to each other, solve for x, and then plug x back into either equation to solve for y. For this problem you don't have to plug the x value back into one of the equations because the problem doesn't ask for the y value.
A. Equation 1: 3x - 2y =6 (subtract 3x from both sides) -2y =-3x +6 (divide both sides by -2) y = 3/2 x -3
Equation 2: 2x+y=4 (subtract 2x from both sides) y = -2x + 4.
Set the equations = to each other: 3/2 x - 3 = -2x + 4 (add 2x to both sides, add 3 to both sides) 7/2x = 7 (multiply both sides by 2/7) x = 2.
A is an answer because the x value of the intersection point is 2.
B: Equation 1: -3x-2y=6 ---> -2y = 3x + 6 ---> y = -3/2 x - 3
Eq 2: -2x+y=4 ---> y = 2x + 4
Set the equations = and solve for x. -3/2 x - 3 = 2x + 4 ---> 1/2 x = 7 ---> x = 14
B isn't an answer.
C: EQ 1 3x + 2y=6 ---> 2y = -3x + 6 ---> y = -3/2 x + 3
EQ 2 : 2x-y= 4 ---> -y = -2x + 4 ---> y = 2x - 4
Set them equal: -3/2 x + 3 = 2x - 4 ---> -7/2 x = - 7 ---> x = 2
C is an answer.
D: EQ1: 3x - 2y =-6 ---> -2y = -3x -6 ---> y = 3/2 x + 3
EQ2: 2x+y = -4 ---> y = -2x - 4
Set the equations = to each other: -2x - 4 = 3/2 x + 3 ---> -3x = 7 ----> x = -7/3 not an answer.
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