Hi

 two lines intersecting at the positive 2 on the x axis.

That would mean the two lines both have x-intercepts = 2

That is... when y = 0  ⇒  x = 2

A. 

3x - 2y =6   || y = 0 ⇒ x = 2

2x + y  =4   || y = 0 ⇒ x = 2 

C. 

3x + 2y=6   || y = 0 ⇒ x = 2 

2x -  y= 4  || y = 0 ⇒ x = 2 

Answer by fcabanski(1391)   (Show Source): You can put this solution on YOUR website!
 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.

Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.) 

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