solve-in-x-R-3x-4-x-5-

Question Number 120953 by bemath last updated on 04/Nov/20

solve in x \in R

∣ 3 x - 4 ∣ = x - 5

Answered by TANMAY PANACEA last updated on 04/Nov/20

w h e n (3 x - 4) > 0 \to x > \frac{4}{3}

(3 x - 4) = x - 5

2 x = - 1

x = \frac{- 1}{5}

b u t x = \frac{- 1}{5} d o e s n o t f o l l o w c o n d i t i o n

x > \frac{4}{3} s o x = \frac{- 1}{5} c a n n o t b e a s o l u t i o n

★ n o w

w h e n (3 x - 4) < 0

x < \frac{4}{3}

- (3 x - 4) = x - 5

- 3 x - x = - 9

4 x = 9

x = \frac{9}{4}

b u t x = \frac{9}{4} d o e s n o t f o l l o w c o n d i t i o n x < \frac{4}{3}

S o n o s o l u t i o n

Answered by ebi last updated on 04/Nov/20

∣ 3 x - 4 ∣= x - 5

C a s e A

3 x - 4 ⩾ 0

3 x - 4 = x - 5

2 x = - 1 \to x = - 0.5

∣ 3 (- 0.5) - 4 ∣ \overset{?}{=} (- 0.5) - 5

∣ - 5.5 ∣ \overset{?}{=} - 5.5

5.5 \neq - 5.5

∴ x = - 0.5 i s n o t t h e s o l u t i o n .

C a s e B

3 x - 4 < 0

3 x - 4 = - (x - 5)

4 x = 9 \to x = 2.25

∣ 3 (2.25) - 4 ∣ \overset{?}{=} 2.25 - 5

∣ - 2.75 ∣ \overset{?}{=} - 2.75

2.75 \neq - 2.75

∴ x = 2.25 i s n o t t h e s o l u t i o n .

n o s o l u t i o n

Commented by MJS_new last updated on 04/Nov/20

∣ 3 \times 2.25 - 4 ∣\neq 2.25 - 5

2.75 \neq - 2.75

Commented by ebi last updated on 04/Nov/20

Y e s . T h a n k y o u s i r f o r n o t i c e t h e m i s t a k e .

Answered by liberty last updated on 04/Nov/20

∣ 3 x - 4 ∣ = x - 5, since ∣ 3 x - 4 ∣ ⩾ 0 for all x ϵ R

then it follow that x ⩾ 5

consider ∣ 3 x - 4 ∣ = x - 5 \to {\begin{cases} 3 x - 4 = x - 5 \\ or \\ 3 x - 4 = 5 - x \end{cases}

{\begin{cases} 2 x = - 1 \to x = - \frac{1}{2} (rejected) \\ 4 x = 9 \to x = \frac{9}{4} (rejected) \end{cases}

thus the equality has no solution

or x = \emptyset .

Answered by mathmax by abdo last updated on 04/Nov/20

∣ 3 x - 4 ∣= x - 5 \Rightarrow∣ 3 x - 4 ∣ - x + 5 = 0 \Rightarrow f (x) = 0

x - \infty \frac{4}{3} + \infty

∣ 3 x - 4 ∣ - 3 x + 4 0 3 x - 4

f (x) - 4 x + 9 2 x + 1

3 x - 4 - x + 5 = 2 x + 1 \Rightarrow f (x) = {\begin{cases} - 4 x + 9 if x ⩽ \frac{4}{3} \\ 2 x + 1 if x ⩾ \frac{4}{3} \end{cases}

if x ⩽ \frac{4}{3} f (x) = 0 \Rightarrow - 4 x + 9 = 0 \Rightarrow x = \frac{9}{4} but \frac{9}{4} > \frac{4}{3} its not solution

if x ⩾ \frac{4}{3} f (x) = 0 \Rightarrow 2 x + 1 = 0 \Rightarrow x = - \frac{1}{2} < \frac{4}{3} \to not solution

so no solution for this equation