To solve the compound inequality \(6b - 1 \leq 213\) and \(2b + 1 \geq 11\), we can solve each part separately.
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Solve \(6b - 1 \leq 213\):
\[ 6b - 1 \leq 213 \]
Add 1 to both sides:
\[ 6b \leq 214 \]
Now, divide both sides by 6:
\[ b \leq \frac{214}{6} \]
Simplify:
\[ b \leq \frac{107}{3} \quad \text{(approximately 35.67)} \]
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Solve \(2b + 1 \geq 11\):
\[ 2b + 1 \geq 11 \]
Subtract 1 from both sides:
\[ 2b \geq 10 \]
Now, divide both sides by 2:
\[ b \geq 5 \]
Now we have the two inequalities:
\[ 5 \leq b \leq \frac{107}{3} \]
Thus, the solution to the compound inequality is:
\[ b \in [5, \frac{107}{3}] \]
or in interval notation,
\[ [5, \frac{107}{3}] \]
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