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I understand how they evaluated the value of x but confuses that why they reject the solution $\frac{-3}{5}$!!!

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  • $\begingroup$ $|x+3|$ is a non-negative number, but when $x=-\frac{3}{5}$ the RHS is negative... $\endgroup$ – fretty Sep 23 '13 at 19:43
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Well, test it. Let $x = -3/5$. Then

$$|x+3| = |-3/5+3| = |12/5| = 12/5 \color{red}{\ne} -12/5 = 4x.$$

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Well, $x$ must be greater or equal to 0, because $4x$ is a module of a number.

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Given that $|x + 3| = 4x$, this implies that $4x$ must be nonnegative (since $|x + 3|$ is clearly nonnegative).

Therefore, $x = -\frac{3}{5}$ is not a solution since $4x$ would then be $-\frac{12}{5}$ and thus negative.

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The absolute value is always non negative. If $x=\frac{-3}{5}$ you would get $0\leq |\frac{-3}{5}+3|=-\frac{12}{5}$ and this is impossible.

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