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Snickers
 5 years ago
solve the system y=ax+b, y=absolute value cx+d
Snickers
 5 years ago
solve the system y=ax+b, y=absolute value cx+d

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should identify the fact that \[y=cx+d=\sqrt{(cx+d)^2}\] by definition.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just wait  I want to see if there's an easier way...but this way will work.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll just keep going. From the definition above, expand the radicand and square both sides to get rid of the square root. Then\[y^2=c^2x^2+2cdx+d^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, square both sides of the other equation,\[y^2=a^2x^2+2abx+b^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0scrap that  I always default to that method...easier to do the following...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Casebycase method: \[y=cx+d=cx+d _.or_.(cx+d)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then, you go about it like you would with other linear equations. You want to find all x such that the same y is yielded. This occurs when,\[ax+b=cx+d_.or_.ax+b=cxd\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solving the first:\[x(ac)=db \rightarrow x=\frac{db}{ac}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and solving the second,\[x(a+c)=db \rightarrow x= \frac{b+d}{a+c}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome...feel free to 'fan' me ;)
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