Problem 36
Question
Find the solution sets of the given inequalities. $$ |x+2|<1 $$
Step-by-Step Solution
Verified Answer
The solution set is \((-3, -1)\).
1Step 1: Understand Absolute Inequality
We are given the inequality \(|x+2| < 1\). This means that the distance from \(x+2\) to 0 is less than 1 on the number line. In simple terms, \(x+2\) lies between -1 and 1.
2Step 2: Rewrite the Compound Inequality
The inequality \(|x+2| < 1\) can be rewritten as the compound inequality: \[-1 < x+2 < 1.\] We are able to rewrite it this way because the absolute value function represents distance, meaning the expression inside the absolute value must lie between the negative and positive of the given number.
3Step 3: Solve for x
Next, we solve the compound inequality for \(x\). Start by isolating \(x\): 1. From \(-1 < x+2\), subtract 2 from all sides: \[-1 - 2 < x < 1 - 2,\] which simplifies to \(-3 < x < -1.\)Thus, the solution set in interval notation for \(x\) is \((-3, -1)\).
Key Concepts
Compound InequalitiesInterval NotationSolving Inequalities
Compound Inequalities
In mathematics, compound inequalities involve two distinct inequalities joined by the words "and" or "or". They help define the range of values that satisfy a condition, allowing us to describe complex relationships using simpler conditions.
When dealing with an inequality like \(|x+2| < 1\), we break it down into two inequalities:
The term "compound" signifies a combination, in this case, combining two linear inequalities. This makes it easier to understand and solve problems where a variable must satisfy several conditions at once. Remember, solving compound inequalities usually involves treating the two inequalities separately at first and later combining their results.
When dealing with an inequality like \(|x+2| < 1\), we break it down into two inequalities:
- \(-1 < x+2 < 1\)
The term "compound" signifies a combination, in this case, combining two linear inequalities. This makes it easier to understand and solve problems where a variable must satisfy several conditions at once. Remember, solving compound inequalities usually involves treating the two inequalities separately at first and later combining their results.
Interval Notation
Interval notation is a shorthand that expresses the set of solutions for inequalities, helping to communicate solutions clearly and concisely.
It uses parentheses and brackets to describe intervals of values:
Interval notation is useful because it provides a clear and standardized way to express solutions, making it easier to communicate with others and ensuring that solution sets are easily understood.
It uses parentheses and brackets to describe intervals of values:
- Parentheses \(( )\) indicate values are not included (open interval).
- Brackets \([ ]\) denote that values are included (closed interval).
Interval notation is useful because it provides a clear and standardized way to express solutions, making it easier to communicate with others and ensuring that solution sets are easily understood.
Solving Inequalities
Solving inequalities is similar to solving equations, but with additional rules to consider, especially when involving absolute values and compound forms.
When solving \(|x+2| < 1\) for \(x\), it is key to think about the absolute value as a distance from zero.
To get rid of the absolute value, rewrite it as a compound inequality:
This results in \(x\) falling between \(-3\) and \(-1\). Always remember that when solving inequalities, the direction of the inequality sign matters, especially when multiplying or dividing by negative numbers—though that's not applied here, it's a crucial tip for inequality solutions in general.
When solving \(|x+2| < 1\) for \(x\), it is key to think about the absolute value as a distance from zero.
To get rid of the absolute value, rewrite it as a compound inequality:
- Divide the inequality into \(-1 < x+2 < 1\).
- From \(-1 < x+2\), get \(-3 < x\).
- From \(x+2 < 1\), get \(x < -1\).
This results in \(x\) falling between \(-3\) and \(-1\). Always remember that when solving inequalities, the direction of the inequality sign matters, especially when multiplying or dividing by negative numbers—though that's not applied here, it's a crucial tip for inequality solutions in general.
Other exercises in this chapter
Problem 36
A right triangle has a fixed hypotenuse of length \(h\) and one leg that has length \(x\). Find a formula for the area \(A(x)\) of the triangle.
View solution Problem 36
Draw the graphs of $$ y=\arcsin x \quad \text { and } \quad y=\arctan \left(x / \sqrt{1-x^{2}}\right) $$ using the same axes. Make a conjecture. Prove it.
View solution Problem 36
Change each rational number to a decimal by performing long division. \(\frac{11}{13}\)
View solution Problem 37
In Problems 35-38, find the slope and \(y\)-intercept of each line. \(6-2 y=10 x-2\)
View solution