To solve the inequality \( 48 \frac{2}{3} \leq -18 + \frac{1}{3}x \), we'll first convert the mixed number \( 48 \frac{2}{3} \) into an improper fraction or a decimal for easier calculations.
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Convert \( 48 \frac{2}{3} \) to an improper fraction: \[ 48 \frac{2}{3} = 48 + \frac{2}{3} = \frac{144}{3} + \frac{2}{3} = \frac{146}{3} \]
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Rewrite the inequality: \[ \frac{146}{3} \leq -18 + \frac{1}{3}x \]
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Add \( 18 \) to both sides of the inequality. First, we convert \( -18 \) into a fraction with a denominator of \( 3 \): \[ -18 = -\frac{54}{3} \] So, adding it gives: \[ \frac{146}{3} + \frac{54}{3} \leq \frac{1}{3}x \] Simplifying the left side: \[ \frac{146 + 54}{3} \leq \frac{1}{3}x \] \[ \frac{200}{3} \leq \frac{1}{3}x \]
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To eliminate the fraction \( \frac{1}{3} \) multiplying \( x \), multiply both sides of the inequality by \( 3 \): \[ 200 \leq x \] This can also be written as: \[ x \geq 200 \]
Thus, the solution to the inequality is: \[ \boxed{x \geq 200} \]
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