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Chapter 0: Problem 91

Solve each absolute value inequality. $$4+\left|3-\frac{x}{3}\right| \geq 9$$

Short Answer

Expert verified
The solution to the inequality is \( x \leq -6 \cup x \geq 24 \)

Step by step solution

01

Isolate the absolute value

Here, we will start by transferring the constant term on the LHS of our inequality to the RHS to isolate the absolute value term. We thus get \( \left|3-\frac{x}{3}\right| \geq 9 - 4 \), which simplifies to \( \left|3-\frac{x}{3}\right| \geq 5 \).
02

Create two separate inequalities

Next, we will form two separate inequalities to take into account the positive and negative values of the expression within the absolute value. This results in \(3-\frac{x}{3} \geq 5 \) for the positive case and \(3-\frac{x}{3} \leq -5 \) for the negative case. Note that in the negative case, the direction of the inequality changes.
03

Solve both inequalities

Now, we solve both inequalities formed in the previous step. For the positive case: \(3-\frac{x}{3} \geq 5 \) simplifies to \( x \leq -6 \). For the negative case: \(3-\frac{x}{3} \leq -5\) simplifies to \( x \geq 24 \).
04

Combine the solutions

Considering both the positive and negative case solutions, though they give us different ranges, this being a 'greater than or equal to' inequality, there will be no values of x that satisfy both inequalities. Hence we have to take union of the solutions from Step 3.

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Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. An elevator at a construction site has a maximum capacity of 2800 pounds. If the elevator operator weighs 265 pounds and each cement bag weighs 65 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

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