Problem 20 - Precalculus Student Solutions Manual 5th

Question

Solve each inequality. Write each solution set in interval notation. $$2-4 x+5(x-1)<-6(x-2)$$

Step-by-Step Solution

Answer

The solution set is \((-\infty, \frac{15}{7})\).

Step 1: Distribute and combine like terms

First, distribute the constants across the variables in the inequality. For the left side: \(2 - 4x + 5(x - 1)\). Distribute 5 across \((x - 1)\) to get \(2 - 4x + 5x - 5\). Simplify by combining like terms to get \(x - 3\). For the right side: \(-6(x - 2)\), distribute -6 across \((x - 2)\) to get \(-6x + 12\). Thus, the inequality becomes \(x - 3 < -6x + 12\).

Step 2: Isolate the variable

Next, add \(6x\) to both sides to get the variable on one side: \(x + 6x - 3 < -6x + 6x + 12\). This simplifies to \(7x - 3 < 12\).

Step 3: Solve for the variable

Add 3 to both sides: \(7x - 3 + 3 < 12 + 3\). This simplifies to \(7x < 15\). Finally, divide both sides by 7: \(x < \frac{15}{7}\). The solution in interval notation is \((-\infty, \frac{15}{7})\).

Key Concepts

interval notation

Interval notation is a way to represent a set of numbers between two endpoints. It's especially useful for expressing solutions to inequalities. Let's look at how it works:

In interval notation, parentheses \( ( , ) \) are used for excluding endpoints, while brackets \( [ , ] \) include them. Here's an example:

• Open intervals: \( (a, b) \) means all values between \( a \) and \( b \), excluding \( a \) and \( b \).
• Closed intervals: \( [a, b] \) includes both \( a \) and \( b \).

In the context of our exercise, the solution \( x < \frac{15}{7} \) is written in interval notation as \( (-\infty, \frac{15}{7}) \). This means it includes all numbers less than \( \frac{15}{7} \) but not \( \frac{15}{7} \) itself.

distributive property

The distributive property is a fundamental algebraic property used to multiply a single term across a sum or difference inside parentheses. It tells us that \( a(b + c) = ab + ac \). In the context of our inequality problem:

• On the left side: \( 2 - 4x + 5(x - 1) \), distribute \( 5 \) over \( x-1 \) to get \( 2 - 4x + 5x - 5 \).
• On the right side: \( -6(x - 2) \), distribute \( -6 \) over \( x-2 \) to get \( -6x + 12 \).

This property helps simplify expressions and is crucial for solving equations and inequalities.

combining like terms

Combining like terms involves simplifying an expression by merging terms with the same variable part. It's an essential step to make equations or inequalities manageable. Let's break it down using our example:

• On the left side: \( 2 - 4x + 5x - 5 \), we combine \( -4x \) and \( 5x \) to get \( x \). So the expression becomes \( x - 3 \).
• On the right side: We already have \( -6x + 12 \) which needs no further simplification.

This results in the simplified inequality \( x - 3 < -6x + 12 \). Combining like terms is all about grouping similar variables and constant terms together.

isolating the variable

Isolating the variable means rearranging the equation or inequality so the variable appears by itself on one side. This step helps us find the solution. Here's how it works in our problem:

• Start with the simplified inequality: \( x - 3 < -6x + 12 \).
• Add \( 6x \) to both sides to move all \( x \)'s to one side: \( x + 6x - 3 < -6x + 6x + 12 \), which simplifies to \( 7x - 3 < 12 \).
• Next, add \( 3 \) to both sides: \( 7x - 3 + 3 < 12 + 3 \), simplifying to \( 7x < 15 \).
• Finally, divide each side by \( 7 \): \( x < \frac{15}{7} \).

This process of isolating the variable allows us to clearly determine the solution to the inequality.