Problem 22 - Algebra and Trigonometry with Analytic Geometry

Question

\(\left\\{\begin{array}{l}|x+2| \leq 1 \\ |y-3|<5\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \(-3 \leq x \leq -1\) and \(-2 < y < 8\).

Step 1: Solve the First Inequality

First, let's solve the inequality \(|x+2| \leq 1\). This absolute value inequality can be split into two separate inequalities:1. \(x+2 \leq 1\)2. \(-(x+2) \leq 1\) or equivalently \(x+2 \geq -1\).Solving these:- From \(x+2 \leq 1\), subtract 2 from both sides to get \(x \leq -1\).- From \(x+2 \geq -1\), subtract 2 from both sides to get \(x \geq -3\).Thus, \(-3 \leq x \leq -1\).

Step 2: Solve the Second Inequality

Now, solve the inequality \(|y-3| < 5\). This also breaks into two separate inequalities:1. \(y - 3 < 5\)2. \(-(y-3) < 5\) or equivalently \(y - 3 > -5\).Solving these:- From \(y - 3 < 5\), add 3 to both sides to get \(y < 8\).- From \(y - 3 > -5\), add 3 to both sides to get \(y > -2\).Thus, the solution for \(y\) is \(-2 < y < 8\).

Step 3: Combine the Solutions

Now, combine the results from the two inequalities.For \(x\), the solution is \(-3 \leq x \leq -1\).For \(y\), the solution is \(-2 < y < 8\).Therefore, the solution set for the system of inequalities is given by the intervals:\(\begin{cases} -3 \leq x \leq -1 \ -2 < y < 8 \end{cases}\)

Key Concepts

Absolute Value Inequality

Absolute value inequalities are a fascinating mathematical concept that help us describe the range of values within a certain distance from a specific point. If you're working with an absolute value inequality, such as \(|x+2| \leq 1\), it essentially means you're looking for all the values of \(x\) that are at most 1 unit away from \(-2\).To solve this, you split it into two separate inequalities:

• \(x+2 \leq 1\)
• \(-(x+2) \leq 1\)

This is because absolute value expressions can be thought of as measuring distance, thus they can "stretch" into two directions — one negative and one positive.When you solve \(x+2 \leq 1\), you simply subtract 2 from both sides, leading to \(x \leq -1\). Similarly, solving \(-(x+2) \leq 1\) simplifies to \(x \geq -3\). By combining these two results, you find that \(-3 \leq x \leq -1\), capturing exactly the values of \(x\) that satisfy this condition.

Interval Notation

Interval notation is a nifty shorthand that lets us communicate solutions to inequalities without writing out all the individual numbers they include. It's a powerful tool for clearly and efficiently stating which numbers make an inequality true.Let's recount our previous results. We found that for \(|x+2| \leq 1\), the solution set is \(-3 \leq x \leq -1\). Similarly, for \(|y-3| < 5\), the solution is \(-2 < y < 8\).In interval notation, these solutions are expressed as:

• For \(x\): \([-3, -1]\), where square brackets [ ] indicate that the endpoints are included (i.e., \(-3\) and \(-1\) are part of the solution set).
• For \(y\): \((-2, 8)\), where parentheses ( ) signify that the endpoints are not included (i.e., \(-2\) and \(8\) are not part of the solution set).

This notation is compact and precise, making it a staple in presenting solutions to systems of inequalities.

Solving Inequalities

Solving inequalities is like navigating through a labyrinth—except here, we find the set of all possible solutions that make our inequalities true. To solve inequalities, we follow certain rules similar to solving equations but with extra care when it comes to operations like multiplying or dividing by negative numbers, as these can flip inequality signs.Start by taking our system of inequalities:

• \(|x+2| \leq 1\)
• \(|y-3| < 5\)

We approach each separately, unraveling absolute value inequalities into two distinct cases. For each case, we solve using standard algebraic manipulations such as adding, subtracting, dividing, and multiplying. For example, if you multiply or divide by a negative number, remember to reverse the inequality symbol.Once solved separately, combine the results, ensuring that each variable's solution respects all of its corresponding inequalities. Hence, for our case, combining \(-3 \leq x \leq -1\) and \(-2 < y < 8\), we state the final solution set as intervals.