Problem 26 - Algebra and Trigonometry

Question

Solve the linear inequality. Express the solution using interval notation and graph the solution set. $$ 2(7 x-3) \leq 12 x+16 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, 11]\). Graph: a filled circle at 11, line shaded left.

Step 1: Expand the Left Side

Start by expanding the left side of the inequality \(2(7x - 3)\). Distribute the 2 to both terms inside the parentheses: \(2 \times 7x - 2 \times 3 = 14x - 6\). The inequality becomes:\[14x - 6 \leq 12x + 16\]

Step 2: Move Variables to One Side

To isolate the variable on one side, subtract \(12x\) from both sides of the inequality. This will give you: \(14x - 12x - 6 \leq 16\) which simplifies to \(2x - 6 \leq 16\).

Step 3: Isolate the Variable Term

Add 6 to both sides of the inequality to further isolate the variable term. This results in: \(2x - 6 + 6 \leq 16 + 6\) or \(2x \leq 22\).

Step 4: Solve for the Variable

Divide both sides of the inequality by 2 to solve for \(x\). This results in: \(x \leq 11\).

Step 5: Express the Solution Set

The inequality \(x \leq 11\) in interval notation is expressed as \(( -\infty, 11 ] \).

Step 6: Graph the Solution Set

To graph the solution set, draw a number line. Mark a filled circle at 11 to indicate that 11 is included in the solution (due to the \(\leq\) sign). Shade the line to the left of 11 to represent all numbers less than or equal to 11.

Key Concepts

Interval Notation

Interval notation is a method of representing a set of numbers, where the numbers lie within specific bounds. It's a type of shorthand that helps to easily convey a range of values on a number line.
In this system, brackets and parentheses are used to indicate whether endpoints are included or excluded in the set.- A square bracket \([\ \ \ ]\) means the endpoint is included. This is also known as a closed interval. For example, \[5, 12\] means that the set includes all numbers from 5 to 12, including 5 and 12.- A parenthesis \(\ (\ \ \ )\) means the endpoint is not included. This is an open interval. For instance, \((-\infty, 11)\) includes all numbers less than 11 but not 11 itself.When solving linear inequalities, like in our exercise where we find the solution is \(x \leq 11\), the interval notation \((-\infty, 11]\) indicates all numbers less than or equal to 11. The infinity symbol \(\infty\) is always paired with a parenthesis because infinity itself is a concept, not a number that can be reached or included.

Graphing Inequalities

Graphing inequalities involves representing the solution set of an inequality on a number line. This visual representation helps to easily understand which numbers are part of the solution and which are not.
Let's break down the process:- Begin by drawing a horizontal line, which serves as your number line.- Identify the critical point or points from the solution set (in our case, it is 11).- Decide on the type of point to use: a filled (solid) circle indicates the endpoint is included, while an open circle indicates it is not. For \(x \leq 11\), use a filled circle at 11.- Shade to the left of the number 11 to cover all values of \(x\) that satisfy the inequality. Shading to the left shows all numbers less than or equal to 11 are part of the solution.This graphical method is intuitive because it visually covers the entire range of solutions while clearly showing which endpoint is included.

Solution Set

The solution set of an inequality is the collection of values that satisfy the inequality. It describes all possible numbers that can make the inequality true. Understanding this helps to know the scope of possible solutions.
There are key elements to consider:- **Boundary Points**: These are specific numbers that mark the transition between satisfying and not satisfying the inequality. In \(x \leq 11\), 11 is a boundary point.- **Range**: This covers the extent of numbers in the solution set. Often written in interval notation, it could extend to \(\infty\) or \(-\infty\) if open-ended.For the inequality \(x \leq 11\), our solution set includes all real numbers less than or equal to 11. This comprehensive approach confirms that any number within this range will consistently satisfy the initial inequality given in the exercise.