Please-guide-me-Sir-I-was-trying-to-solve-this-eq-for-searching-possible-values-of-x-eq-is-x-2-lt-3-x-7-the-range-of-x-whom-i-got-23-2-lt-x-lt-19

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 by maxmathsup 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 by afachri 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 by afachri last updated on 11/Dec/18

i^{'} m glad to be part of this forum .

Hail Math!

Commented by Abdo 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 \dots

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 by afachri 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 by hassentimol last updated on 11/Dec/18

You are welcome sir . It was a pleasure .

Commented by hassentimol last updated on 11/Dec/18

Well \dots any number smaller than - 19 / 4 will

verify the equation .

But a number which would be greater than

2 would also satisfy the inequation \dots

Commented by hassentimol last updated on 11/Dec/18

Commented by hassentimol 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 \dots

Commented by afachri 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 by afachri last updated on 11/Dec/18

Commented by afachri 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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