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Question Number 49857 by afachri last updated on 11/Dec/18

$Please guide me Sir . I was trying to solve$ $this eq for searching possible values of x .$ $eq is :$ $∣ x - 2 ∣ < 3 ∣ x + 7 ∣$ $the range of x whom i got : - \frac{23}{2} < x < - \frac{19}{4}$ $but the result do not satisfy the eq, instead$ $i put x > - 4, they satisfy the eq .$ $please help me out of this pickle .$ $Not because i {didn}^{'} t try, yet i always$ $stuck in this type of function .$
Commented bymaxmathsup by imad last updated on 11/Dec/18

$t h e b e s t w a y i s t h i s$ $l e t A (x) =∣ x - 2 ∣ - 3 ∣ x + 7 ∣ (e) \Leftrightarrow A (x) < 0 w e e r a d i c a t e t h e a b s l u t e v a l u e$ $x - \infty - 7 2 + \infty$ $∣ x - 2 ∣ - x + 2 - x + 2 0 x - 2$ $∣ x + 7 ∣ - x - 7 0 x + 7 x + 7$ $A (x) 2 x + 23 - 4 x - 19 - 2 x - 23$ $c a s e 1 x ⩽ - 7 (e) \Leftrightarrow 2 x + 23 < 0 \Leftrightarrow x < - \frac{23}{2} \Rightarrow S_{1} =] - \infty, - \frac{23}{2} [$ $c a s e 2 - 7 ⩽ x ⩽ 2 (e) \Leftrightarrow - 4 x - 19 < 0 \Leftrightarrow - 4 x < 19 \Leftrightarrow 4 x > - 19 \Leftrightarrow x > - \frac{19}{4}$ $\Rightarrow S_{2} =] - \frac{19}{4}, 2]$ $c a s e 3 x ⩾ 2 \Rightarrow (e) \Leftrightarrow - 2 x - 23 < 0 \Leftrightarrow - 2 x < 23 \Leftrightarrow 2 x > - 23 \Leftrightarrow x > - \frac{23}{2}$ $\Rightarrow S_{3} = [2, + \infty [a n d w e t a k e \cup S_{i}$ $S = S_{1} \cup S_{2} \cup S_{3} =] - \infty, - \frac{23}{2} [\cup] - \frac{19}{4}, + \infty [.$
Commented byafachri last updated on 11/Dec/18

$yes it is Sir . Seems this be my another references .$ $thank you Mr Max . soon after Mr Hassen was$ $giving me explanation, i re - tried and get$ $the answer - \frac{23}{2} > x > - \frac{19}{4} . and i found$ $match answer to both of yours Sir .$
Commented byafachri last updated on 11/Dec/18

$i^{'} m glad to be part of this forum .$ $Hail Math!$
Commented byAbdo msup. last updated on 12/Dec/18

$y o u a r e w e l c o m e$
	Answered by hassentimol last updated on 11/Dec/18
	$Well... We write it as : ∣x−2∣ − 3∣x+7∣ < 0 We have : { ((x−2 > 0 ⇔ x > 2)),((x+7 > 0 ⇔ x > −7)) :} Therefore we define I_1 , I_2 and I_3 such as : I_1 = ]−∞,−7] : −x+2+3x+21 = 2x+23 I_2 =[−7,2] : −x+2−3x−21 = −4x−19 I_3 =[2,+∞[ : x−2−3x−21 = −2x−23 We have : Let S_n be the set of solutions for the I_n interval. For I_1 : 2x+23<0 ⇔ x<((−23)/2) : S_1 =]−∞,((−23)/2)[ For I_2 : −4x−19<0 ⇔ x>((−19)/4) : S_2 =[((−19)/4),2[ For I_3 : −2x−23<0 ⇔ x>((−23)/2) : S_3 =[2,+∞[ Finally : We can consider S the set of solutions : S = { x ∣ x∈]−∞, ((−19)/4)[ ∪ ]2_ ^ , +∞[ } In other terms : Either x<((−19)/4) , either x>2. Thanks T.H.$
	$Well . . .$ $We write it as : ∣ x - 2 ∣ - 3 ∣ x + 7 ∣ < 0$ $We have : {\begin{cases} x - 2 > 0 \Leftrightarrow x > 2 \\ x + 7 > 0 \Leftrightarrow x > - 7 \end{cases}$ $Therefore we define I_{1}, I_{2} and I_{3} such as :$ $I_{1} =] - \infty, - 7] : - x + 2 + 3 x + 21 = 2 x + 23$ $I_{2} = [- 7, 2] : - x + 2 - 3 x - 21 = - 4 x - 19$ $I_{3} = [2, + \infty [: x - 2 - 3 x - 21 = - 2 x - 23$ $We have :$ $L e t S_{n} b e t h e s e t o f s o l u t i o n s f o r t h e I_{n} i n t e r v a l .$ $For I_{1} : 2 x + 23 < 0 \Leftrightarrow x < \frac{- 23}{2} : S_{1} =] - \infty, \frac{- 23}{2} [$ $For I_{2} : - 4 x - 19 < 0 \Leftrightarrow x > \frac{- 19}{4} : S_{2} = [\frac{- 19}{4}, 2 [$ $For I_{3} : - 2 x - 23 < 0 \Leftrightarrow x > \frac{- 23}{2} : S_{3} = [2, + \infty [$ $Finally :$ $We can consider S the set of solutions :$ $S = {x ∣ x \in] - \infty, \frac{- 19}{4} [\cup] \underset{}{\overset{}{2}}, + \infty [}$ $In other terms :$ $Either x < \frac{- 19}{4}, either x > 2 .$ $T h a n k s$ $T . H .$
	Commented byafachri last updated on 11/Dec/18

	$Thanks for your explanation Mr . Hassen .$ $I will learn from this .$ $But Sir, is x < - \frac{19}{4} satisfy the eq ? ? ?$
	Commented byhassentimol last updated on 11/Dec/18

	$You are welcome sir . It was a pleasure .$
	Commented byhassentimol last updated on 11/Dec/18

	$Well . . . any number smaller than - 19 / 4 will$ $verify the equation .$ $But a number which would be greater than$ $2 would also satisfy the inequation . . .$
	Commented byhassentimol last updated on 11/Dec/18

	Commented byhassentimol last updated on 11/Dec/18

	$This is the curve of the function when all is put$ $on one side of the inequation .$ $The solutions are when the curve is negative . . .$
	Commented byafachri last updated on 11/Dec/18

	$i put x = (- 5) for trial,$ $∣ x - 2 ∣ - 3 ∣ x + 7 ∣ < 0$ $∣ - 5 - 2 ∣ - 3 ∣ - 5 + 7 ∣ < 0$ $7 - 3 (2) < 0$ $why x = (- 5) {doesn}^{'} t satisfy the eq Sir ?$ $correct me Sir, please .$
	Commented byafachri last updated on 11/Dec/18

	Commented byafachri last updated on 11/Dec/18

	$thank you very much Sir . i^{'} ve looked in$ $my graphic too and understand now .$ $:)$
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