#1**0 **

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. | ||

| Now, solve both equations. | ||

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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