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Snickers

  • 5 years ago

solve the system y=ax+b, y=absolute value cx+d

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  1. anonymous
    • 5 years ago
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    You should identify the fact that \[y=|cx+d|=\sqrt{(cx+d)^2}\] by definition.

  2. anonymous
    • 5 years ago
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    Just wait - I want to see if there's an easier way...but this way will work.

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

  4. anonymous
    • 5 years ago
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    Now, square both sides of the other equation,\[y^2=a^2x^2+2abx+b^2\]

  5. anonymous
    • 5 years ago
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    scrap that - I always default to that method...easier to do the following...

  6. anonymous
    • 5 years ago
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    Case-by-case method: \[y=|cx+d|=cx+d _.or_.-(cx+d)\]

  7. anonymous
    • 5 years ago
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    Then, 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,\[ax+b=cx+d_.or_.ax+b=-cx-d\]

  8. anonymous
    • 5 years ago
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    Solving the first:\[x(a-c)=d-b \rightarrow x=\frac{d-b}{a-c}\]

  9. anonymous
    • 5 years ago
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    and solving the second,\[x(a+c)=-d-b \rightarrow x= -\frac{b+d}{a+c}\]

  10. Snickers
    • 5 years ago
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    cool thank you.

  11. anonymous
    • 5 years ago
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    You're welcome...feel free to 'fan' me ;)

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spraguer (Moderator)
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