Question 179219
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You must have it mixed up with something else that you probably should forget anyway.  If you don't want <i>x</i> to be negative, then you have to do something to it to make it not negative.  But if you do anything to one side of an equation, then you must do the same thing to the other side in order to maintain equality between the two expressions. 


If *[tex \Large -x = 9] then you need to multiply both sides of the equation by -1 to get *[tex \Large (-1)(-x) = (-1)(9) \rightarrow x = -9] 


As to your description of your process with *[tex \Large -x=-8y+9], I think you have things backwards.  Generally, <i>x</i> is the independent variable, and  <i>y</i> is the dependent variable.  That means that you should be selecting values for <i>x</i> and then solving for the value of <i>y</i> that depends upon that selection.  Your first step should be to solve (rearrange) the equation so that you have the value of <i>y</i> in terms of <i>x</i>, thus:


*[tex \Large -x=-8y+9]


Add 8y and x to both sides:


*[tex \Large -x + x + 8y = -8y + 8y + x + 9]


*[tex \Large 8y = x + 9]


Multiply both sides by *[tex \LARGE \frac {1}{8}]


*[tex \Large \left(\frac {1}{8}\right) 8y = \left(\frac {1}{8}\right) (x + 9)]


*[tex \Large y = \left(\frac {1}{8}\right) (x + 9)]


Now you can select values for <i>x</i> and determine the resulting value for <i>y</i>.


Let *[tex \Large x = 1 \rightarrow y = \frac {10}{8} = \frac {5}{4}]


Therefore the ordered pair *[tex \Large (x, y) = (1, \frac {5}{4})] is one point on the line.


Let *[tex \Large x = -1 \rightarrow y = \frac {8}{8} = 1]


Therefore the ordered pair *[tex \Large (x, y) = (-1, 1)] is a second point on the line giving you sufficient information to uniquely determine, and therefore graph, the line by plotting the two points and drawing the line that contains both of them.


{{{drawing(
600, 600, -5, 5, -5, 5,
grid(1),
circle(1,5/4,.1),
circle(-1,1,.1),
graph(
600, 600, -5, 5, -5, 5,
y = (1/8)(x + 9)
))}}}



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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