Question - Solving Absolute Value Inequality
Solution:
To solve the inequality $$ 3|1 - 9b| \geq 28 $$, you can follow these steps:1. Divide both sides of the inequality by 3 to isolate the absolute value expression:$$ |1 - 9b| \geq \frac{28}{3} $$2. Recognize that an absolute value expression $$ |x| \geq a $$ where $$ a > 0 $$ can be split into two separate inequalities:$$ x \geq a $$ or $$ x \leq -a $$In this case, $$ x $$ corresponds to $$ 1 - 9b $$ and $$ a $$ corresponds to $$ \frac{28}{3} $$.3. Set up the two inequalities:$$ 1 - 9b \geq \frac{28}{3} $$ and $$ 1 - 9b \leq -\frac{28}{3} $$4. Solve each inequality separately:For $$ 1 - 9b \geq \frac{28}{3} $$:a) Subtract 1 from both sides:$$ -9b \geq \frac{28}{3} - 1 $$$$ -9b \geq \frac{28}{3} - \frac{3}{3} $$$$ -9b \geq \frac{25}{3} $$b) Divide by -9, remembering to flip the inequality sign when dividing by a negative:$$ b \leq -\frac{25}{27} $$For $$ 1 - 9b \leq -\frac{28}{3} $$:a) Subtract 1 from both sides:$$ -9b \leq -\frac{28}{3} - 1 $$$$ -9b \leq -\frac{28}{3} - \frac{3}{3} $$$$ -9b \leq -\frac{31}{3} $$b) Divide by -9, flipping the inequality sign:$$ b \geq \frac{31}{27} $$5. Combine the two parts of the solution:$$ b \leq -\frac{25}{27} $$ or $$ b \geq \frac{31}{27} $$This is the solution to the inequality.
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