Could you please help me for this question : Solve for x : ∣ 3x + 2 ∣ + ∣ 7x − 5 ∣ = 20 Thank you


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Question Number 47224 by hassentimol last updated on 06/Nov/18

$Could you please help me for this question :$ $Solve for x :$ $∣ 3 x + 2 ∣ + ∣ 7 x - 5 ∣ = 20$ $Thank you$
Commented by maxmathsup by imad last updated on 06/Nov/18

$l e t f i r s t e r a d i c t e t h e a b s o l u t v a l u e$ $x - \infty - \frac{2}{3} \frac{5}{7} + \infty$ $∣ 3 x + 2 ∣ - 3 x - 2 0 3 x + 2 3 x + 2$ $∣ 7 x - 5 ∣ - 7 x + 5 - 7 x + 5 0 7 x + 5$ $A (x) - 10 x + 3 - 4 x + 7 10 x + 7$ $e q u . \Leftrightarrow A (x) = 20$ $c a s e 1 x \in] - \infty, - \frac{2}{3}] A (x) = 20 \Leftrightarrow - 10 x + 3 = 20 \Leftrightarrow - 10 x = 17 \Leftrightarrow x = - \frac{17}{10}$ $- \frac{17}{10} + \frac{2}{3} = \frac{- 34 + 20}{30} = \frac{- 14}{30} < 0 \Rightarrow - \frac{17}{10} < - \frac{2}{3} s o t h e n u m b e r i s s o l u t i o n$ $c a s e 2 x \in [- \frac{2}{3}, \frac{5}{7}] A (x) = 20 \Leftrightarrow - 4 x + 7 = 20 \Leftrightarrow - 4 x = 13 \Leftrightarrow x = - \frac{13}{4}$ $- \frac{13}{4} + \frac{2}{3} < 0 \Rightarrow - \frac{13}{4} < - \frac{2}{3} s o t h i s n u m b e r i s n t s o l u t i o n$ $c a s e 3 x \in [\frac{5}{7}, + \infty [A (x) = 20 \Leftrightarrow 10 x + 7 = 20 \Leftrightarrow 10 x = 13 \Leftrightarrow x = \frac{13}{10}$ $\frac{13}{10} - \frac{5}{7} = \frac{13.7 - 50}{70} > 0 \Rightarrow \frac{13}{10} > \frac{5}{7} s o t h i s n u m b e r i s s o l u t i o n s o t h e (e) h a v e$ $2 s o l u t i o n s - \frac{17}{10} a n d \frac{5}{7} .$ $∣$
Commented by hassentimol last updated on 07/Nov/18

$Thank you sir!$ $I^{'} ve understood thanks to you .$
Commented by maxmathsup by imad last updated on 07/Nov/18

$y o u a r e w e l c o m e . .$
	Answered by MJS last updated on 06/Nov/18

	$x < - \frac{2}{3} \Rightarrow ∣ 3 x + 2 ∣= - 3 x - 2$ $x ⩾ - \frac{2}{3} \Rightarrow ∣ 3 x + 2 ∣= 3 x + 2$ $x < \frac{5}{7} \Rightarrow ∣ 7 x - 5 ∣= 5 - 7 x$ $x ⩾ \frac{5}{7} \Rightarrow ∣ 7 x - 5 ∣= 7 x - 5$ $x \in I_{1} =] - \infty; - \frac{2}{3} [$ $- 3 x - 2 + 5 - 7 x = 20 \Rightarrow x = - \frac{17}{10}$ $x \in I_{2} = [- \frac{2}{3}; \frac{5}{7} [$ $3 x + 2 + 5 - 7 x = 20 \Rightarrow x = - \frac{13}{4} \notin I_{2} \Rightarrow no solution$ $x \in I_{3} = [\frac{5}{7}; + \infty [$ $3 x + 2 + 7 x - 5 = 20 \Rightarrow x = \frac{23}{10}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18

	$c r i t i c a l v a l u e o f x a r e \frac{- 2}{3} a n d \frac{5}{7}$ $\frac{- 2}{3} \approx - 0.67 s n d \frac{5}{7} \approx 0.71$ $w h e n x > o . 71$ $∣ 3 x + 2 ∣= 3 x + 2$ $∣ 7 x - 5 ∣= 7 x - 5$ $3 x + 2 + 7 x - 5 = 20$ $10 x = 23$ $x = \frac{23}{10} = 2.3$ $w h e n 0.71 > x > - 0.67 (s a y x = 0)$ $∣ 3 x + 2 ∣= 3 x + 2$ $∣ 7 x - 5 ∣= - (7 x - 5)$ $3 x + 2 - 7 x + 5 = 20$ $- 4 x = 13$ $x = \frac{- 13}{4} = - 3.25$ $b u t - 3.25 < - 0.67$ $s o x c a n n o t b e - 3.25$ $w h e n x < - 0.67 (s a y x = - 1)$ $∣ 3 x + 2 ∣= - (3 x + 2)$ $∣ 7 x - 5 ∣= - (7 x - 5)$ $- 3 x - 2 - 7 x + 5 = 20$ $- 10 x = 17$ $x = - 1.7$ $s o s o l u t i o n$ $p l s c h e c k . . . .$ $y e s l e t m e r e c t i f y . . .$ $2.3 a n d - 1.7 o v e r l o o k e d = s i g n$
	Commented by hassentimol last updated on 06/Nov/18

	$Thank you Sir,$ $God bless you$
	Commented by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

	$t o m a k e o n e u n d e r s t a n d i s m y g i f t . . .$
	Commented by hassentimol last updated on 06/Nov/18

	$May I ask you a question ?$ $Why would \frac{- 13}{4} = - 3.25 may not be one$ $of the answers if it is smaller than 0.67 ?$ $I know that there can be only two answers$ $as drawm on a graph, but why is this$ $solution incorrect instead of the other ones ?$ $Than you .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18

	$t h a n k y o u s i r . . .$
	Answered by peter frank last updated on 07/Nov/18

	$for positive$ $(3 x + 2) + (7 x - 5) = 20$ $10 x - 3 = 20$ $10 x = 23$ $x = 2.3$ $for - ve$ $- (3 x + 2) + - (7 x - 5)$ $- 3 x - 2 - 7 x + 5 = 20$ $- 10 x + 3 = 20$ $x = - 1.7$
	Commented by maxmathsup by imad last updated on 06/Nov/18

	$s i r p e t e r t h i s i s n o t t h e w a y t o s o l v e e q u a t i o n l i k e t h i s i i n v i e y o u t o u s e$ $m y m e t h o d g i v e n a t t h e a n s w e r . . .$
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