Problem 52 - A Graphical Approach to College Algebra

Question

Solve each equation or inequality. $$|3(x-5)+2|+3=9$$

Step-by-Step Solution

Answer

Solutions: \(x = \frac{19}{3}\) and \(x = \frac{7}{3}\).

Step 1: Isolate the Absolute Value

First, we need to isolate the absolute value in the equation. Start by subtracting 3 from both sides of the equation: $$|3(x-5)+2|+3 = 9$$ becomes $$|3(x-5)+2| = 6.$$

Step 2: Set Up Two Cases for the Absolute Value

The expression inside the absolute value can be either positive or negative, leading to two possible cases: 1. \(3(x-5) + 2 = 6\)2. \(3(x-5) + 2 = -6\).

Step 3: Solve the First Case

For the equation \(3(x-5) + 2 = 6\):1. Subtract 2 from both sides: \(3(x-5) = 4\).2. Divide by 3: \(x - 5 = \frac{4}{3}\).3. Add 5 to both sides to solve for \(x\): \(x = \frac{4}{3} + 5\), which simplifies to \(x = \frac{19}{3}\).

Step 4: Solve the Second Case

For the equation \(3(x-5) + 2 = -6\):1. Subtract 2 from both sides: \(3(x-5) = -8\).2. Divide by 3: \(x - 5 = \frac{-8}{3}\).3. Add 5 to both sides to solve for \(x\): \(x = \frac{-8}{3} + 5\), which simplifies to \(x = \frac{7}{3}\).

Step 5: Write the Solution

The solutions to the equation are \(x = \frac{19}{3}\) and \(x = \frac{7}{3}\). Both values satisfy the initial absolute value equation.

Key Concepts

Solving Equations

To tackle any absolute value equation, the first step is to isolate the absolute value expression. In the problem given, this means shifting all other terms away from the absolute value term. Once isolated, you can transform the problem into two separate scenarios because an absolute value represents the distance from zero, which can be either positive or negative.

After isolation, divide the problem into two linear equations based on the positive and negative scenarios. For example:

• Equation when the expression is positive: set the inside expression equal to the positive number.
• Equation when the expression is negative: set the inside expression equal to the negative of that number.

Solve each of these equations individually by performing algebraic operations like addition, subtraction, multiplication, or division. In this context, you're aiming to isolate the variable on one side of the equation. Successfully solving both equations will provide the final solutions to the absolute value equation.

Algebraic Expressions

Understanding and manipulating algebraic expressions is crucial when solving equations, especially those involving absolute values. These expressions are combinations of numbers, variables, and arithmetic operations. When dealing with absolute value equations, focus on simplifying these expressions to make solving for the variable manageable.

For example, in the equation provided, you have the expression inside the absolute value \(3(x-5) + 2\).
You'll need to expand and simplify it:

• First, distribute any numbers multiplied by a group: multiply 3 to both \(x\) and \(-5\) which gives you \(3x - 15\).
• Add the 2 to this expression so your simplified form becomes \(3x - 15 + 2\) or \(3x - 13\).

These simplifications set the stage to apply the absolute value solutions effectively. By mastering these concepts, algebraic expression handling becomes systematic and clear.

Inequalities

Inequalities show the relationship of variable values when they are not equal and can be expressed as less than or greater than. Absolute value inequalities are a unique subset that you may encounter, where the solution represents a range of values.

The basic principle involves handling the equation very similarly to absolute value equations—split the absolute inequality into two possible conditions that together depict the range of solutions:

• The positive scenario, where the expression is greater than or equal to the positive number.
• The negative scenario, where the expression is less than or equal to the negative number.

Solving these inequalities involves finding solution sets, often resulting in intervals that satisfy the inequality.
Understanding these concepts can aid in grasping the full range of possible solution methods and enhances algebraic skills beyond equations, into areas involving greater analysis of mathematical scenarios.