1/3I4p-11I=p+4

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.