This is the equation:
$|3x+6|>-12$.
I solved it under two cases ($3x+6>-12$) and ($3x+6<12$).
More than the answer, what I need to know is -
Have I rightly constructed those $2$ cases, if wrong, please explain.
Thanks.
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jameselmore
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Romy
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4You don't need to set the two cases up, because the solution set can be easily seen to be all real numbers. – RK01 Jul 22 '15 at 13:03
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yes it is $|x|\geq 0$ but in our inequality we have $|3x+6|>-12$ – Dr. Sonnhard Graubner Jul 22 '15 at 13:06
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OP: Did you already draw the graph of the function $x\mapsto|x|$? What does it look like? – Did Jul 22 '15 at 13:09
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@Did - No sir, i am not that good at Mathematics, i was following this [link]http://www.mhhe.com/math/precalc/barnettpc5/graphics/barnett05pcfg/ch01/others/bpc5_ch01-04.pdf . And theorem 3 is what precisely i was looking into. – Romy Jul 22 '15 at 13:13
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What do they say just below "Definition 1" already? (And note that "Theorem 3" assumes that p>0 hence applying it blindly for p=-12 can only lead to disasters.) – Did Jul 22 '15 at 13:23
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@Did - I got it now, won't make this mistake again :d Also, i would like to mention that Wolframalpha says 'All values of x are solutions' but Mathway says 'No Solution'. – Romy Jul 22 '15 at 13:35
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Yeah, just remember that programs like WA and Mathway are good at what they do and inept at what they cannnot do. – Did Jul 22 '15 at 13:52
2 Answers
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For any complex/real number $a,$ $$|a|\ge0$$ which greater than any negative number
lab bhattacharjee
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Note that $|3x+6|$ is always non-negative. So, $|3x+6|\gt -12$ holds for every $x\in\mathbb R$.
For $|ax+b|\gt -c$ in the title,
If $c=0$, then it always holds except $x=-\frac ba$ (for $a\not=0$).
If $-c\lt 0$, then it always holds. (your case is included here)
If $-c\gt 0$, we have $$|ax+b|\gt -c\iff ax+b\lt -(-c)\ \ \ \text{or}\ \ \ ax+b\gt -c$$
For $C\gt 0$, we have
$$|x|\gt C\iff x\lt -C\ \ \ \text{or}\ \ \ x\gt C$$ $$|x|\lt C\iff -C\lt x\lt C$$
mathlove
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