moongazer
  • moongazer
Teach me in solving this inequality.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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moongazer
  • moongazer
|2x−11|−2>4x+7
moongazer
  • moongazer
I know how to solve equations. But I just don't know how to solve if it has absolute value sign. Teach me please. :)
anonymous
  • anonymous
x < 1/3

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moongazer
  • moongazer
how?
moongazer
  • moongazer
Teach me please and give me another example after teaching it. so that I can practice. and you will going to check it if I am right.
anonymous
  • anonymous
To remove the modulus sign we write \[|2x -11|\] as \[\sqrt{2x - 11}^2\]
moongazer
  • moongazer
Why is that?
vishal_kothari
  • vishal_kothari
http://www.mathmotivation.com/lectures/Inequalities.pdf
anonymous
  • anonymous
To remove any modulus sign write \[\left| ax + b \right| = (\sqrt{ax + b})^2\]
moongazer
  • moongazer
Could you solve it step by step? and what is the reason behind To remove any modulus sign write
moongazer
  • moongazer
\[\left| ax + b \right| = (\sqrt{ax + b})^2\]
hoblos
  • hoblos
|2x−11|−2>4x+7 |2x−11| > 4x +9 then 2x-11 > 4x +9 OR 2x-11 < -(4x+9) -2x > 20 2x -11 < -4x -9 x<-10 6x < 2 x<1/3
moongazer
  • moongazer
please explain
hoblos
  • hoblos
rules: if |x| < a then -aa then x>a OR x<-a
hoblos
  • hoblos
you apply the second rule here
moongazer
  • moongazer
Thanks!
moongazer
  • moongazer
What if it is "="
hoblos
  • hoblos
if |x| = a then x=a OR x=-a

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