Snickers
  • Snickers
solve the system y=ax+b, y=absolute value cx+d
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
You should identify the fact that \[y=|cx+d|=\sqrt{(cx+d)^2}\] by definition.
anonymous
  • anonymous
Just wait - I want to see if there's an easier way...but this way will work.
anonymous
  • anonymous
I'll just keep going. From the definition above, expand the radicand and square both sides to get rid of the square root. Then\[y^2=c^2x^2+2cdx+d^2\]

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anonymous
  • anonymous
Now, square both sides of the other equation,\[y^2=a^2x^2+2abx+b^2\]
anonymous
  • anonymous
scrap that - I always default to that method...easier to do the following...
anonymous
  • anonymous
Case-by-case method: \[y=|cx+d|=cx+d _.or_.-(cx+d)\]
anonymous
  • anonymous
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,\[ax+b=cx+d_.or_.ax+b=-cx-d\]
anonymous
  • anonymous
Solving the first:\[x(a-c)=d-b \rightarrow x=\frac{d-b}{a-c}\]
anonymous
  • anonymous
and solving the second,\[x(a+c)=-d-b \rightarrow x= -\frac{b+d}{a+c}\]
Snickers
  • Snickers
cool thank you.
anonymous
  • anonymous
You're welcome...feel free to 'fan' me ;)

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