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Applying the definition of module and simplifications, fill in the gaps. |π‘₯| βˆ’ 2|π‘₯ βˆ’ 3| = { π‘₯ βˆ’ 6, 𝑠𝑒 π‘₯ < 0 ___________, 𝑠𝑒 0 ≀ π‘₯ < ______ ______, 𝑠𝑒_________________

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Answer to a math question Applying the definition of module and simplifications, fill in the gaps. |π‘₯| βˆ’ 2|π‘₯ βˆ’ 3| = { π‘₯ βˆ’ 6, 𝑠𝑒 π‘₯ < 0 ___________, 𝑠𝑒 0 ≀ π‘₯ < ______ ______, 𝑠𝑒_________________

Expert avatar
Rasheed
4.7
110 Answers
|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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