Problem 18
Question
Find the values of \(x\) that satisfy the inequalities. $$ x+1>2 \text { or } x-1<-2 $$
Step-by-Step Solution
Verified Answer
The values of x that satisfy the given inequalities are \(x < -1 \text{ or } x > 1\).
1Step 1: Solve each inequality separately
We have two inequalities, let's solve them one by one.
For inequality \(x + 1 > 2\):
To isolate the x, subtract 1 from both sides of the inequality: \(x > 1\)
For inequality \(x - 1 < -2\):
To isolate the x, add 1 to both sides of the inequality: \(x < -1\)
2Step 2: Combine the solutions
Since the original inequality has the "or" sign, any value of x that satisfies either of the inequalities will be a solution.
For inequality \(x > 1\), the x values greater than 1 are the solutions.
For inequality \(x < -1\), the x values less than -1 are the solutions.
So the solution of the given compound inequality is:
\(x < -1 \text{ or } x > 1\)
The values of x that satisfy the given inequalities are the values less than -1 or greater than 1.
Key Concepts
Compound InequalitiesSolving InequalitiesMathematical Solutions
Compound Inequalities
Compound inequalities are a special type of mathematical statement that involves more than one inequality being solved at the same time. Unlike a simple inequality, which involves only one statement, compound inequalities combine multiple inequalities using conjunctions like "and" or "or".
In a statement involving "and," both conditions must be true at the same time. Conversely, an "or" compound inequality requires that at least one of the conditions be true to satisfy the inequality.
It's useful to see compound inequalities as having a broader solution set since they provide multiple scenarios in which the statement holds true. Understanding compound inequalities is crucial when solving more complex mathematical problems, as they often form the backbone of mathematical reasoning.
In a statement involving "and," both conditions must be true at the same time. Conversely, an "or" compound inequality requires that at least one of the conditions be true to satisfy the inequality.
It's useful to see compound inequalities as having a broader solution set since they provide multiple scenarios in which the statement holds true. Understanding compound inequalities is crucial when solving more complex mathematical problems, as they often form the backbone of mathematical reasoning.
Solving Inequalities
Solving inequalities is quite similar to solving equations, with some key differences. The goal is to isolate the variable, usually represented by an "x", on one side of the inequality sign (>, <, ≥, ≤).
To solve an inequality, you need to perform operations that maintain the inequality's truth – such as adding, subtracting, multiplying, or dividing both sides by the same number. However, a crucial point to remember is that when you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign.
In our example of the first inequality \(x+1 > 2\):
To solve an inequality, you need to perform operations that maintain the inequality's truth – such as adding, subtracting, multiplying, or dividing both sides by the same number. However, a crucial point to remember is that when you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign.
In our example of the first inequality \(x+1 > 2\):
- Subtract 1 from both sides to isolate \(x\), yielding \(x>1\).
- Add 1 to both sides to get \(x < -1\).
Mathematical Solutions
Mathematical solutions to inequalities reveal the set of possible values that satisfy the given statements. For compound inequalities that include "or," such as \(x+1 > 2\ or\ x-1 < -2\), the solution consists of values that satisfy either condition. This broadens the possible values that \(x\) can take.
For instance, if you solve \(x+1 > 2\), you find that \(x > 1\). On the other hand, solving \(x-1 < -2\) gives you \(x < -1\).
For instance, if you solve \(x+1 > 2\), you find that \(x > 1\). On the other hand, solving \(x-1 < -2\) gives you \(x < -1\).
- The solution is all values of \(x\) such that \(x > 1\) or \(x < -1\).
Other exercises in this chapter
Problem 18
Perform the indicated operations and simplify. \(\frac{4 y+12}{y+2} \cdot \frac{3 y+6}{2 y-1}\)
View solution Problem 18
Solve the equation by completing the square. $$ x^{2}-x-6=0 $$
View solution Problem 18
Rewrite the number without radicals or exponents. $$ 81^{-1 / 4} $$
View solution Problem 18
Solve the given equation. $$ \frac{1}{3}[2-3(x+2)]=\frac{1}{4}\left[(-3 x+1)+\frac{1}{2} x\right] $$
View solution