How do you solve #|(x/2) + 3| = 3x -3#?
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To solve the equation |(x/2) + 3| = 3x - 3, follow these steps:
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Split the equation into two cases: Case 1: (x/2) + 3 = 3x - 3 Case 2: (x/2) + 3 = -(3x - 3)
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Solve each case separately for x.
Case 1: (x/2) + 3 = 3x - 3 Solve for x: x/2 + 3 = 3x - 3 x/2 - 3x = -3 - 3 x/2 - 3x = -6 x - 6x = -12 -5x = -12 x = (-12)/(-5) x = 12/5
Case 2: (x/2) + 3 = -(3x - 3) Solve for x: (x/2) + 3 = -3x + 3 x/2 + 3x = 3 - 3 x/2 + 3x = 0 Multiply both sides by 2 to eliminate the fraction: x + 6x = 0 7x = 0 x = 0
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Check the solutions: Substitute each solution back into the original equation to ensure they are valid.
For x = 12/5: |(12/5)/2 + 3| = 3(12/5) - 3 |6/5 + 3| = 36/5 - 3 |33/5| = 36/5 - 3 33/5 = 36/5 - 15/5 33/5 = 21/5 33 ≠ 21, so this solution is not valid.
For x = 0: |0/2 + 3| = 3(0) - 3 |3| = 0 - 3 3 = -3 This is not true, so the solution x = 0 is also not valid.
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Therefore, the equation has no real solutions.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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