SOLUTION: The following are from the Compass sample test. Algebra Placement Linear equations 15.What is the slope of the line with the equation 2x+3y+6=0 Please explain how to get to

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Question 101398: The following are from the Compass sample test.
Algebra Placement
Linear equations
15.What is the slope of the line with the equation 2x+3y+6=0
Please explain how to get to the answer of -2/3
16. Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of point B so that the line x=2 is the perpendicular bisector of AB?
Please explain how to get to the answer of (8,1)

Answer by doukungfoo(195) About Me  (Show Source):
You can put this solution on YOUR website!
15.What is the slope of the line with the equation 2x+3y+6=0

To find the slope we need to convert this equation to the slope intercept form which is:
y=mx+b
m is the slope and b is y intercept.
2x + 3y + 6 = 0
first move 2x to the right side of the equation
This is done by subtracting 2x from both sides
2x - 2x + 3y + 6 = 0-2x
on the left side 2x-2x cancel out and on the right side 0-2x leaves -2x so we get this
3y + 6 = -2x
next move 6 following the same procedure
3y + 6 - 6 = -2x - 6
3y = -2x - 6
finally isolate y by dividing by 3 across the equation
%283y%2F3%29=-%282%2F3%29x-%286%2F3%29
y=-%282%2F3%29x-2%29
Now that we have convert the equation to slope intercept form we can identify the slope, which is:
-2%2F3%29
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16. Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of point B so that the line x=2 is the perpendicular bisector of AB?

For this one I would just plot the information given on a graph.
First plot point A at (-4,1) Then draw the line x=2 which is a vertical line that crosses the x axis at 2. Once we have done this we can see that point A is 6 units away from the the line x=2 so point B must be 6 units away on the x axis for line x=2 to be a bisector of line segement AB. Also point B must have the same x intercept as point A for line segment AB to be perpendicular to line x=2. The only point on the graph that meets these requirements is (8,1)

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