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# 1/3I4p-11I=p+4

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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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