Problem 43
Question
Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$2|2 x+1|+1=19$$
Step-by-Step Solution
Verified Answer
The solutions are \(x = 4\) and \(x = -5\). Interval notation: \{-5, 4\}.
1Step 1: Simplify the equation
First, isolate the absolute value expression on one side of the equation. Start by subtracting 1 from both sides of the equation: \[2|2x+1| + 1 - 1 = 19 - 1\]This simplifies to:\[2|2x+1| = 18\]
2Step 2: Divide both sides by 2
Next, divide both sides of the equation by 2 to solve for the absolute value:\[\frac{2|2x+1|}{2} = \frac{18}{2}\]This results in:\[|2x+1| = 9\]
3Step 3: Set up the equations for absolute value
The absolute value equation \(|2x+1| = 9\) means we need to consider two equations: one where the expression inside the absolute value is positive and one where it is negative. This gives us:1. \(2x+1 = 9\)2. \(2x+1 = -9\)
4Step 4: Solve for x in the positive case
For the first equation, solve for \(x\):\[2x + 1 = 9\]Subtract 1 from both sides:\[2x = 8\]Now, divide both sides by 2:\[x = 4\]
5Step 5: Solve for x in the negative case
For the second equation, solve for \(x\):\[2x + 1 = -9\]Subtract 1 from both sides:\[2x = -10\]Now, divide both sides by 2:\[x = -5\]
6Step 6: Express the solution in interval notation
The solutions for \(x\) are \(x = 4\) and \(x = -5\). In interval notation, this does not create a true interval since these are discrete solutions. Thus, the solution set in interval format is expressed as:\[\{-5, 4\}\]
Key Concepts
Interval NotationSolving EquationsAlgebraic ExpressionsDiscrete Solutions
Interval Notation
Interval notation is a way of describing sets of numbers, particularly intervals on the number line. It uses brackets and parentheses to indicate whether endpoints are included or not. In general,
- Use square brackets [ ] to include an endpoint in the interval, known as a closed interval.
- Use parentheses ( ) to exclude an endpoint, known as an open interval.
- Combine brackets and parentheses for half-open intervals.
Solving Equations
Solving equations is about finding values for variables that make the equation true. The basic idea is to isolate the variable you are solving for. Here's a brief overview of the common steps:
- Start by simplifying the equation if necessary. This means removing parentheses or combining similar terms.
- Use operations like addition, subtraction, multiplication, or division to isolate the variable on one side of the equation.
- Check your solution by substituting the value back into the original equation.
Algebraic Expressions
Algebraic expressions are a combination of numbers, variables, and operations. They can range from simple expressions like \(2x+1\) to more complex ones. In the context of solving equations, these are transformed and manipulated to isolate the variable:
- Identify the expression that needs to be simplified.
- Apply arithmetic operations to both sides equally to maintain the equality.
- Ensure the equation remains balanced by following the same operation on both sides.
Discrete Solutions
Not all solutions form continuous intervals; some are discrete. Discrete solutions are individual values where the equation holds true. These can often be represented in set notation:
- Each solution is a unique point on the number line.
- They do not occur in a continuous sequence.
- Examples include counts of items, specific integer solutions, or points like \(\{-5, 4\}\).
Other exercises in this chapter
Problem 43
Approximate the real-number expression. Express the answer in sclentific notation accurate to four significant figures. (a) \(\frac{1.2 \times 10^{3}}{3.1 \time
View solution Problem 43
Find the real solutions of the equation. (a) \(x^{5 / 3}=32\) (b) \(x^{4 / 3}=16\) (c) \(x^{2 / 3}=-36\) (d) \(x^{34}=125\) (e) \(x^{3 / 2}=-27\)
View solution Problem 43
Simplify the expression. $$\frac{2 x}{x+2}-\frac{8}{x^{2}+2 x}+\frac{3}{x}$$
View solution Problem 43
Find the solutions of the equation. $$x^{2}-5 x+20=0$$
View solution