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Mathematics 8 Online
OpenStudy (anonymous):

How can ax+by+cz=d be written as f(x,y)=Ax+By+D?

OpenStudy (anonymous):

By expressing z In this case z=(d-ax-by)/c therefore f(x,y)=(d-ax-by)/c

OpenStudy (anonymous):

But how's that equal to Ax+By+D?

OpenStudy (anonymous):

Well that's what it says so here: http://tutorial.math.lamar.edu/Classes/CalcIII/MultiVrbleFcns.aspx

OpenStudy (zarkon):

\[f(x,y)=\frac{d-ax-by}{c}=\frac{-ax-by+d}{c}=\frac{-a}{c}x+\frac{-b}{c}y+\frac{d}{c}\] \[=Ax+By+D\] where \[A=\frac{-a}{c},B=\frac{-b}{c},D=\frac{d}{c}\]

OpenStudy (anonymous):

Are there books or resources that show this same derivation as well? Particularly about the part saying that\[A=\frac{-a}{c}, B=\frac{-b}{c}, D=\frac d c\]

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