Problem 37
Question
Solve each linear inequality in Exercises 27-48 and graph the solution set on a number line. Express the solution set using interval notation. $$2 x-11<-3(x+2)$$
Step-by-Step Solution
Verified Answer
The solution to the inequality \(2x - 11 < -3x -6\) is \(x < 1\). The solution in interval notation is \(-\infty , 1\).
1Step 1: Distribute
Distribute -3 to both \(x\) and 2 in \(-3(x+2)\), making the inequality \(2x - 11 < -3x -6\).
2Step 2: Rearrange
Add 3x to both sides to gather all terms containing x on the left side of the inequality and simplify. The equation now is \(5x - 11 < -6\).
3Step 3: Solve for x
Add 11 to both sides to isolate x on one side of the inequality. The equation now is \(5x < 5\). Now, divide both sides of the inequality by 5 to solve for x. So, the solution is \(x < 1\).
4Step 4: Number Line and Interval Notation
On a number line, an open circle is drawn at 1 and an arrow is drawn to the left, as x is less than 1. In interval notation, the solution is expressed as \(-\infty, 1\).
Key Concepts
Inequality GraphingInterval NotationAlgebraic Manipulation
Inequality Graphing
Understanding the graphing of inequalities is crucial for visualizing the range of solutions in a given problem. To graph an inequality, you begin by treating the inequality as a strict equality to identify critical points. If the inequality is strictly 'less than' < or 'greater than' >, a dot called an open circle is placed on the number line indicating that this point is not included in the solution set. However, if the inequality is 'less than or equal to' ≤ or 'greater than or equal to' ≥, a closed circle is used instead to show that the value is part of the solution set.
In our example, the inequality was resolved to be \(x < 1\). On the number line, we place an open circle at 1 and draw an arrow extending to the left, which signifies all values less than 1. This visual representation quickly communicates that any number to the left of the open circle is a valid solution to the inequality. It's important when graphing inequalities to ensure the direction of the arrow corresponds with the inequality sign: to the left for 'less than' and to the right for 'greater than'.
In our example, the inequality was resolved to be \(x < 1\). On the number line, we place an open circle at 1 and draw an arrow extending to the left, which signifies all values less than 1. This visual representation quickly communicates that any number to the left of the open circle is a valid solution to the inequality. It's important when graphing inequalities to ensure the direction of the arrow corresponds with the inequality sign: to the left for 'less than' and to the right for 'greater than'.
Interval Notation
Interval notation is an alternative and concise way to express the set of solutions for an inequality. It uses parentheses, ( or ), to denote open intervals, where endpoints are not included, and brackets, [ or ], for closed intervals, where they are.
In the given problem, since \(x\) is less than 1, but 1 is not included in the solution, we use parentheses to express the solution as \(-\infty, 1\). The \(-\infty\) symbol represents that the solutions can go indefinitely to the left, towards increasingly negative numbers. Interval notation is particularly powerful because it provides a clear, unambiguous method to describe continuous sets of numbers, such as the range of solutions to inequalities.
In the given problem, since \(x\) is less than 1, but 1 is not included in the solution, we use parentheses to express the solution as \(-\infty, 1\). The \(-\infty\) symbol represents that the solutions can go indefinitely to the left, towards increasingly negative numbers. Interval notation is particularly powerful because it provides a clear, unambiguous method to describe continuous sets of numbers, such as the range of solutions to inequalities.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging, simplifying, and solving algebraic expressions and equations. It is the toolkit we employ to isolate the variable of interest and find its range of possible values when dealing with inequalities.
In our practice exercise, we employed algebraic manipulation by first distributing the \(-3\) across the \(x+2\) in \(-3(x+2)\). Next, we rearranged the inequality by collecting like terms and moving all terms containing \(x\) to one side, which often involves using the addition or subtraction of the same quantity from both sides. After that, we isolated \(x\) by dividing or multiplying both sides, as we did by dividing the entire inequality by 5. Each step was careful to maintain the balance of the inequality and use inverse operations to progress towards finding the solution. Effective algebraic manipulation is both an art and a science, requiring a solid understanding of mathematical rules and a strategic approach to solving the puzzle presented by the inequality.
In our practice exercise, we employed algebraic manipulation by first distributing the \(-3\) across the \(x+2\) in \(-3(x+2)\). Next, we rearranged the inequality by collecting like terms and moving all terms containing \(x\) to one side, which often involves using the addition or subtraction of the same quantity from both sides. After that, we isolated \(x\) by dividing or multiplying both sides, as we did by dividing the entire inequality by 5. Each step was careful to maintain the balance of the inequality and use inverse operations to progress towards finding the solution. Effective algebraic manipulation is both an art and a science, requiring a solid understanding of mathematical rules and a strategic approach to solving the puzzle presented by the inequality.
Other exercises in this chapter
Problem 37
Determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then write and factor the trinomial. $$ x^{2}-\frac{1
View solution Problem 37
You are choosing between two plans at a discount warehouse. Plan A offers an annual membership fee of \(\$ 100\) and you pay \(80 \%\) of the manufacturer's rec
View solution Problem 37
Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator ze
View solution Problem 38
In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$\frac{-12+\sqrt{-28}}{32}$$
View solution