# 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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3. 👁 195
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

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

Good job!

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posted by Guido

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