Problem 66 - Algebra and Trigonometry

Question

In Exercises 59–94, solve each absolute value inequality. $$ |3(x-1)+2| \leq 20 $$

Step-by-Step Solution

Answer

The solution to the inequality |3(x-1)+2| ≤ 20 is \( x \in [-6.333, 7] \).

Step 1: Remove the Absolute Value

In order to solve this, we need to first convert the absolute value inequality into a regular inequality. Since |a| ≤ b means -b ≤ a ≤ b, we rewrite |3(x-1)+2| ≤ 20 as -20 ≤ 3(x-1) + 2 ≤ 20.

Step 2: Solve the Inequalities

Now we have two inequalities to solve: -20 ≤ 3(x-1)+2 and 3(x-1)+2 ≤ 20. We simplify these inequalities:\First inequality:\Distribute 3 to get -20 ≤ 3x - 3 + 2. Combine like terms to become -20 ≤ 3x - 1, and then add 1 to both sides to get -19 ≤ 3x. Finally, divide each side by 3 to get -6.333 ≤ x.\Second inequality:\Distribute 3 to get 3x - 3 + 2 ≤ 20. Combine like terms to become 3x - 1 ≤ 20, and then add 1 to both sides to get 3x ≤ 21. Finally, divide each side by 3 to get x ≤ 7.

Step 3: Write Final Solution

The solution to both inequalities is that x is greater than or equal to -6.333 and less than or equal to 7. This can be written in interval notation as [-6.333, 7]. This is the interval of real numbers x for which the inequality |3(x-1)+2| ≤ 20 holds true.

Key Concepts

Algebraic Inequalities

Algebraic inequalities are mathematical expressions where two values are compared by inequality signs such as <, >, ≤, and ≥. Unlike equations, which state that two expressions are equal, inequalities show that there is a difference in value or size, with one side being greater or less than the other.

They can be simple, comparing just two numbers, like 2 < 5, or more complex with variables and expressions, such as the one we see in our example: |3(x-1)+2| ≤ 20. When solving algebraic inequalities involving absolute values, it's important to remember that the absolute value of a number is its distance from zero on the number line, regardless of direction. This means we need to consider both the positive and negative scenarios when removing the absolute value symbol to create two separate inequalities to solve.

Solving Inequalities

The process of solving inequalities is akin to solving equations, but with a special consideration for the direction of the inequality. To solve an inequality, we perform operations that maintain the balance of the inequality, just like equations. However, one important rule to remember is that if you multiply or divide both sides by a negative number, you must flip the inequality sign.

In our example, after removing the absolute value, we have two separate inequalities to solve. Key steps here include distributing any multiplicative factors across terms, combining like terms, and isolating the variable of interest by adding or subtracting quantities to both sides. Dividing both sides by a positive number does not change the direction of the inequality. Through this methodical approach, we can find the range of values that satisfy the initial inequality.

Interval Notation

Interval notation is a mathematical shorthand for expressing a range of numbers. It is especially handy when describing the solutions to inequalities, where there's a set of values that satisfy the inequality's conditions. The notation uses brackets and parentheses to communicate inclusivity or exclusivity of the endpoints.

A bracket, such as [a, b], indicates that the range includes the endpoints 'a' and 'b'. Conversely, a parenthesis, (a, b), shows that 'a' and 'b' are not included in the range. In our exercise, the solution set is shown as \[ -6.333, 7 \], which means all real numbers from -6.333 to 7, including -6.333 and 7 themselves, satisfy the original absolute value inequality.