Problem 82 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation and inequality analytically. Use interval notation to write the solution set for each inequality. (a) \(5-3 x=x+1\) (b) \(5-3 x \leq x+1\) (c) \(5-3 x \geq x+1\)

Step-by-Step Solution

Answer

(a) \(x = 1\); (b) \([1, \infty)\); (c) \((- infty, 1]\).

Step 1: Solve the equation 5 - 3x = x + 1

To solve for \( x \), first, bring all terms involving \( x \) to one side and constant terms to the other. Subtract \( x \) from both sides: \[ 5 - 3x - x = x + 1 - x \] Simplifying, we get: \[ 5 - 4x = 1 \] Next, subtract 5 from both sides: \[ 5 - 4x - 5 = 1 - 5 \] This yields: \[ -4x = -4 \] Finally, divide both sides by \( -4 \): \[ x = 1 \] Thus, the solution to the equation is \( x = 1 \).

Step 2: Solve the inequality 5 - 3x ≤ x + 1

For this inequality, let's isolate \( x \) just like we did with the equation. Start by subtracting \( x \) from both sides:\[ 5 - 3x - x \leq x + 1 - x \] Simplify: \[ 5 - 4x \leq 1 \] Subtract 5 from both sides: \[ 5 - 4x - 5 \leq 1 - 5 \] This gives us: \[ -4x \leq -4 \] Divide by \( -4 \) and remember to flip the inequality sign: \[ x \geq 1 \] In interval notation, the solution set is \([1, \infty)\).

Step 3: Solve the inequality 5 - 3x ≥ x + 1

For the third inequality, similar steps are taken. Start by subtracting \( x \) from both sides:\[ 5 - 3x - x \geq x + 1 - x \] Simplify: \[ 5 - 4x \geq 1 \] Subtract 5 from both sides: \[ 5 - 4x - 5 \geq 1 - 5 \] We get: \[ -4x \geq -4 \] Divide by \( -4 \) and flip the inequality sign: \[ x \leq 1 \] The solution in interval notation is \((-\infty, 1]\).

Key Concepts

Interval Notation

Interval notation is a straightforward way to describe the set of solutions for an inequality. Instead of listing every individual possible solution, interval notation provides a more compact form. It utilizes brackets and parentheses to define the start and end points of these solutions.

In interval notation:

• Square brackets [ ] are used to include an endpoint in the set. For example, \([a, b]\) includes both \(a\) and \(b\).
• Parentheses ( ) are used to exclude an endpoint from the set. For example, \((a, b)\) means neither \(a\) nor \(b\) is included.

When writing solutions to inequalities:

• A solution like \([1, \infty)\) describes all numbers starting from 1 and increasing without bound, including 1 but excluding infinity.
• Similarly, \((-\infty, 1]\) includes all numbers less than or equal to 1.

Interval notation makes it easy to understand the range of possible values that fulfill an inequality.

Linear Equations

Linear equations are equations where the variable appears to the power of one. They are the simplest form of equations and take the form: \( ax + b = 0 \). Solving these equations involves isolating the variable on one side.

Here's how you solve a linear equation similarly to the example in our problem:

• Gather terms with the variable on one side of the equation, like moving \( x \) terms together.
• Move constant terms to the opposite side by subtracting or adding the required values from both sides.
• Finally, solve for the variable by dividing by the coefficient attached to the variable.

For instance, in solving \( 5 - 3x = x + 1 \), all steps lead to isolating \( x \) so we find \( x = 1 \). This approach works for any linear equation, making it a foundational skill in algebra.

Analytical Methods

Analytical methods involve using algebraic techniques to solve equations and inequalities. They don't rely on graphical or numerical approximations, making them exact and reliable. Let's see how these methods work, particularly with inequalities:

For an inequality like \( 5 - 3x \leq x + 1 \), follow these steps:

• First, handle it similarly to an equation by moving terms with \( x \) to one side: \( 5 - 4x \leq 1 \).
• Simplify the expression by subtracting constant terms on the opposite side.
• When dividing or multiplying by a negative number, remember to reverse the inequality sign. This step is crucial to finding the correct solution.

This method is systematic and efficient. Once you isolate the variable, using analytical methods provides the exact intervals, as seen when converting \( x \geq 1 \) to \([1, \infty)\). Analytical techniques form the core of solving equations and inequalities in mathematics, providing clear, precise solutions.