Solve the equation:

\[ 6|6-4x| = 8x + 4 \]

Also, check for any extraneous solutions.

\ \ -4x \geq -6\ \ -->\ \ \ x \leq \frac{3}{2}\\\\ 6(6-4x)=8x+4\\ 36-24x=8x+4\ \ |subtract\ 4\\ 32-24x=8x\ \ | add\ 24x\\ 32=32x\ \ | divide\ by\ 32\\ x=1\ \ \ and\ 1 \leq \frac{3}{2}\\\\ II\\ for\ 6-4x <0\ \ \ --> \ \ -4x < -6\ \ -->\ \ \ x > \frac{3}{2}\\\\ 6(-6+4x)=8x+4\\ -36+24x=8x+4\ \ | subtract4\\ -40+24x=8x\ \ \ | subtract\ 24x\\ -40=-16x\ \ | divide\ by\ -16"> 6∣6 − 4 x ∣ = 8 x + 4 F or m u l a f or ab so l u t e v a l u e ∣ a ∣ = a , f or a ≥ 0 or ∣ a ∣ = − a , f or a < 0 I f or 6 − 4 x ≥ 0 − − > − 4 x ≥ − 6 − − > x ≤ 2 3 ​ 6 ( 6 − 4 x ) = 8 x + 4 36 − 24 x = 8 x + 4 ∣ s u b t r a c t 4 32 − 24 x = 8 x ∣ a dd 24 x 32 = 32 x ∣ d i v i d e b y 32 x = 1 an d 1 ≤ 2 3 ​ II f or 6 − 4 x < 0 − − > − 4 x < − 6 − − > x > 2 3 ​ 6 ( − 6 + 4 x ) = 8 x + 4 − 36 + 24 x = 8 x + 4 ∣ s u b t r a c t 4 − 40 + 24 x = 8 x ∣ s u b t r a c t 24 x − 40 = − 16 x ∣ d i v i d e b y − 16 \frac{3}{2}\\\\Solutions:\\ x\in \{1;2,5\}"> x = 2 , 5 an d 2 , 5 > 2 3 ​ S o l u t i o n s : x ∈ { 1 ; 2 , 5 }

Answered by luana | 2024-06-10

The solutions to the equation 6∣6 − 4 x ∣ = 8 x + 4 are x = 1 and x = 2.5 . Both solutions satisfy the original equation, confirming they are valid. Thus, the final solutions are valid and no extraneous solutions were found.
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Answered by luana | 2024-12-26