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21 Absolute Value and Equations

 

\bullet\qquad  |a| = b \quad (b>0) \qquad \Longleftrightarrow\qquad a= b \quad\hbox{ or }\quad a=-b 

ex.

    \[\qquad  |x-5| = 3 \qquad \Longleftrightarrow\qquad x-5= 3 \quad\hbox{ or }\quad x-5=-3\]

    \[\qquad \qquad\Longleftrightarrow\qquad x= 8 \quad\hbox{ or }\quad  x=2\]

 

\bullet\qquad  |a| = |b| \quad\quad \qquad \Longleftrightarrow\qquad a= b \quad\hbox{ or }\quad a=-b

ex.

    \[\qquad  |x-5| = |-x+3| \qquad \Longleftrightarrow\qquad x-5= -x+3 \quad\hbox{ or }\quad x-5=-(-x+3)\]

    \[\qquad \qquad\Longleftrightarrow\qquad (2x= 8) \quad\hbox{ or }\quad (0. x=2  \quad\hbox{impossible})\]

    \[\qquad \qquad\Longleftrightarrow\qquad x=4\]

 

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Exercise 1

 

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The possible solutions are : \qquad \displaystyle{x= -33  \qquad \hbox{or} \qquad  x= 31}.

 

Exercise 2

 

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The possible solutions are : \qquad \displaystyle{x= - \frac{{9}}{{26}}  \qquad \hbox{or} \qquad  x= - \frac{{27}}{{16}} }.

 

Exercise 3

 

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The possible solutions are : \qquad \displaystyle{x=  \frac{{2}}{{28}}   \qquad \hbox{or} \qquad  x=  \frac{{14}}{{28}} }.

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Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.