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How would I solve I=Lines in absolute value equation 1/3I4p-11I=p+4

Guest Sep 6, 2017
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Hello, equations with absolute value bars can be mysterious sometimes, so I'll help you out. The original equation is \(\frac{1}{3}\left|4p-11\right|=p+4\).

 

\(\frac{1}{3}\left|4p-11\right|=p+4\) Multiply by 3 on both sides.
\(|4p-11|=3p+12\) The absolute value symbol can be replaced with a plus-minus symbol, which creates 2 separate equations.
\(\pm(4p-11)=3p+12\) Now, split this up into 2 equations.
\(4p-11=3p+12\) \(-(4p-11)=3p+12\)

 

Now, solve both equations.
\(4p=3p+23\) \(-4p+11=3p+12\)

 

 
\(p=23\) \(-7p=1\)

 

 
\(p_1=23\) \(p_2=-\frac{1}{7}\)

 

 
   

 

These could be extraneous solutions, so we must check to ensure that both are indeed valid solutions.

 

\(\frac{1}{3}\left|4*23-11\right|=23+4\)  
\(\frac{1}{3}\left|81\right|=27\) Multiply by 3 on both sides.
\(|81|=81\)  
\(81=81\) This is true, so \(p_1=23\) is a valid solution.
   

 

Now, let's check \(p=-\frac{1}{7}\):

 

\(\frac{1}{3}\left|4*-\frac{1}{7}-11\right|=-\frac{1}{7}+4\) Do the same simplification as last time.
\(\frac{1}{3}\left|\frac{-4}{7}+\frac{-77}{7}\right|=\frac{-1}{7}+\frac{28}{7}\)  
\(\frac{1}{3}|\frac{-81}{7}|=\frac{27}{7}\) Muliply by 3 on both sides.
\(|\frac{-81}{7}|=\frac{81}{7}\)  
\(\frac{81}{7}=\frac{81}{7}\)  
   

 

Therefore, both solutions are valid solutions.

TheXSquaredFactor  Sep 6, 2017

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