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applying the definition of module and simplifications fill in the gaps x 2 x 3 x 6 se x 0 se 0 x se

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Applying the definition of module and simplifications, fill in the gaps. |𝑥| − 2|𝑥 − 3| = { 𝑥 − 6, 𝑠𝑒 𝑥 < 0 _, 𝑠𝑒 0 ≤ 𝑥 < , 𝑠𝑒_______

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Answer to a math question Applying the definition of module and simplifications, fill in the gaps. |𝑥| − 2|𝑥 − 3| = { 𝑥 − 6, 𝑠𝑒 𝑥 < 0 _, 𝑠𝑒 0 ≤ 𝑥 < , 𝑠𝑒_______

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|x| - 2|x - 3| = \begin{cases} x - 6, & \text{if } x < 0 \\3x - 6, & \text{if } 0 \leq x < 3 \\-x + 6, & \text{if } x \geq 3 \end{cases}

1. For \(x < 0\):

|x| = -x \quad \text{and} \quad |x - 3| = -x + 3

-x - 2(-x + 3) = x - 6

2. For \(0 \leq x < 3\):

|x| = x \quad \text{and} \quad |x - 3| = -x + 3

x - 2(-x + 3) = 3x - 6

3. For \(x \geq 3\):

|x| = x \quad \text{and} \quad |x - 3| = x - 3

x - 2(x - 3) = -x + 6

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