Algebra

1)Find x in the solution of the system 3x+y=2 and 2x-3y=16
A)2
B)-4
C)18/11
D)10/11
I chose A

3x+y=2 times 3 = 9x+3y=6
2x-3y=16 times 1 = 2x-3y=16
11x/11 = 22/11x = 2

2)Find the coordinates of the vertices of the figures formed by y -< x + 2, x + 2 -< 6, and y >- -2
A)(0,0),(2,4),(8,-2)
B)(-4,-2),(2,4),(8,-2)
C)(-4,-2),(4,2),(8,-2)
D)(-2,-4),(2,4),(8,-2)
I chose B

this one confused me this is all the work to show:
-2 -< -4 + 2
2 + 4 -< 6
-2

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asked by Jon
  1. Your inequations form the boundaries of the figure.
    So change the < or > to an equal sign, then solve them in pairs.
    You should get 3 sets of intersection points.
    Solve the pairs of equations just like you did #1.

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    posted by Reiny
  2. ok im stuck after this:
    y = x + 2
    x + y = 6
    y = -2

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    posted by Jon
  3. I will do number one only.

    We have this:

    (1)Find x in the solution of the system 3x + y = 2 and 2x -3y = 16.

    (A)2
    (B)-4
    (C)18/11
    (D)10/11

    Your choice = (A)

    We have a system of linear equations in two variables.

    Here are the two equations:

    3x + y = 2...Equations A
    2x -3y = 16...Equation B

    I will first isolate y in Equation A.

    3x + y = 2

    y = -3x + 2....I will plug the quantity (-3x + 2) in Equation B to find the value of x.

    2x - 3y = 16

    2x - 3(-3x + 2) = 16

    2x + 9x - 6 = 16

    11x - 6 = 16

    11x = 16 + 6

    11x = 22

    x = 22/11

    x = 2

    Your choice is correct!

    Good job!

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    posted by Guido

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