MTH120 College Algebra Chapter 2.7

 2.7 Linear Inequalities and Absolute Value Inequalities

Linear inequalities and absolute value inequalities involve inequalities with linear expressions or absolute value expressions. Solving these types of inequalities is similar to solving equations, but with some differences due to the presence of inequalities. Let's look at both types:

1. Linear Inequalities:

A linear inequality is an inequality involving linear expressions (polynomials of degree 1). The general form of a linear inequality is:

$��+�<�$

where $�$, $�$, and $�$ are constants, and $�$ is not equal to 0. Solving a linear inequality involves finding the values of $�$ that satisfy the inequality.

Steps for Solving Linear Inequalities:

1. Treat the inequality like an equation, but with one difference: if you multiply or divide both sides of the inequality by a negative number, reverse the inequality sign.

2. Solve for $�$ to find the solution set.

3. Graph the solution set on the number line, using open circles for < or > signs and filled circles for ≤ or ≥ signs, and shading in the appropriate direction.

Example: Solve the inequality $2�-3>7$.

1. Treat it like an equation and add 3 to both sides:

   $2�-3+3>7+3$

   $2�>10$

2. Divide both sides by 2:

   $\frac{2�}{2}>\frac{10}{2}$

   $�>5$

3. Graph the solution on the number line:

   

2. Absolute Value Inequalities:

Absolute value inequalities involve absolute value expressions, which are denoted by |x|. The general form of an absolute value inequality is:

$\mathrm{\mid }�(�)\mathrm{\mid }<�$

where $�(�)$ is a function of $�$ and $�$ is a positive constant. Solving an absolute value inequality involves finding the values of $�$ that make the absolute value expression less than a given value $�$.

Steps for Solving Absolute Value Inequalities:

1. Set up two inequalities, one with the expression inside the absolute value bars positively and the other with the expression inside the absolute value bars negatively.

2. Solve both inequalities separately.

3. Combine the solutions from both inequalities.

Example: Solve the absolute value inequality $\mathrm{\mid }2�-3\mathrm{\mid }<5$.

1. Set up two inequalities:

   $2�-3<5$ (positively)

   $-2�+3<5$ (negatively)

2. Solve both inequalities separately:

   a) $2�-3<5$

   Add 3 to both sides:

   $2�<8$

   Divide by 2 (since 2 is positive):

   $�<4$

   b) $-2�+3<5$

   Subtract 3 from both sides:

   $-2�<2$

   Divide by -2 (since -2 is negative, reverse the inequality sign):

   $�>-1$

3. Combine the solutions:

   $-1<�<4$

This means that $�$ can be any value between -1 and 4 (excluding -1 and 4) to satisfy the absolute value inequality.

These are the basics of solving linear inequalities and absolute value inequalities. Depending on the specific inequality, the solution set may vary. Always double-check your solution by substituting values from the solution set back into the original inequality to ensure it's correct.

Interval notation is a concise and convenient way to represent the solutions of inequalities and absolute value inequalities. It uses brackets and parentheses to denote different types of intervals on the real number line. Here's how interval notation works:

1. Closed Interval: A closed interval includes its endpoints and is represented using square brackets [ ]. For example, [a, b] includes all values of x between a and b, including a and b.

2. Open Interval: An open interval does not include its endpoints and is represented using parentheses ( ). For example, (a, b) includes all values of x between a and b but excludes a and b.

3. Half-Open or Half-Closed Interval: These intervals include one endpoint and exclude the other. You can mix square brackets and parentheses to represent these intervals. For example, [a, b) includes all values greater than or equal to a and less than b.

Here are some examples of how to use interval notation:

Example 1: Represent the interval of all real numbers greater than or equal to -3 and less than 5.

Interval Notation: $[-3,5)$

Example 2: Represent the interval of all real numbers greater than -2 but less than 4.

Interval Notation: $(-2,4)$

Example 3: Represent the interval of all real numbers less than or equal to 1.

Interval Notation: $(-\mathrm{\infty },1]$

Example 4: Represent the interval of all real numbers greater than 2.

Interval Notation: $(2,\mathrm{\infty })$

Example 5: Represent the interval of all real numbers greater than or equal to -1 and less than 3.

Interval Notation: $[-1,3)$

Example 6: Represent the interval of all real numbers that are at most 7 but greater than -4.

Interval Notation: $(-4,7]$

Interval notation is especially useful when dealing with complex inequalities and systems of inequalities, as it provides a clear and concise way to express the solution sets on the real number line.

To express all real numbers greater than or equal to a using interval notation, you would use the closed interval notation. Here are some examples:

Example 1: Express all real numbers greater than or equal to 2 using interval notation.

Interval Notation: $[2,\mathrm{\infty })$

This interval includes all real numbers starting from 2 and going to positive infinity. The square bracket at 2 indicates that 2 is included in the interval.

Example 2: Express all real numbers greater than or equal to -1 using interval notation.

Interval Notation: $[-1,\mathrm{\infty })$

This interval includes all real numbers starting from -1 and going to positive infinity. The square bracket at -1 indicates that -1 is included in the interval.

Example 3: Express all real numbers greater than or equal to 0 using interval notation.

Interval Notation: $[0,\mathrm{\infty })$

This interval includes all real numbers starting from 0 and going to positive infinity. The square bracket at 0 indicates that 0 is included in the interval.

In each of these examples, the interval starts at the given number (2, -1, or 0) and continues indefinitely in the positive direction. The square bracket indicates that the endpoint is included in the interval, and the infinity symbol ($\mathrm{\infty }$) represents positive infinity.

Using the properties of inequalities is a fundamental aspect of solving and manipulating inequalities. Here are some common properties and techniques you can use when working with inequalities:

1. Addition and Subtraction Properties:

• Adding or Subtracting the Same Value: You can add or subtract the same value to both sides of an inequality without changing the inequality's direction. For example, if $�<�$, then $�+�<�+�$ for any real number $�$.

2. Multiplication and Division Properties:

• Multiplying or Dividing by a Positive Number: If you multiply or divide both sides of an inequality by a positive number, the direction of the inequality remains unchanged. For example, if $�<�$ and $�>0$, then $��<��$.

• Multiplying or Dividing by a Negative Number: If you multiply or divide both sides of an inequality by a negative number, the direction of the inequality is reversed. For example, if $�<�$ and $�<0$, then $��>��$.

3. Combining Inequalities:

• Adding or Subtracting Inequalities: You can add or subtract two inequalities if they have the same direction. For example, if $�<�$ and $�<�$, then $�+�<�+�$.

• Multiplying or Dividing Inequalities: You can multiply or divide two inequalities if they have the same direction and both sides are positive. For example, if $�<�$ and $�<�$ (with $�,�,�,�>0$), then $��<��$.

4. Absolute Value Inequalities:

• When dealing with absolute value inequalities, you often need to consider both the positive and negative cases separately. For example, to solve $\mathrm{\mid }�\mathrm{\mid }<3$, you would consider two cases: $�<3$ and $-�<3$.

5. Inequality Symbols:

• Less Than (<): This symbol represents "is less than." For example, $�<5$ means $�$ is smaller than 5.

• Greater Than (>): This symbol represents "is greater than." For example, $�>3$ means $�$ is larger than 3.

• Less Than or Equal To ($\le$): This symbol represents "is less than or equal to." For example, $�\le 2$ means $�$ is less than or equal to 2.

• Greater Than or Equal To ($\ge$): This symbol represents "is greater than or equal to." For example, $�\ge 4$ means $�$ is greater than or equal to 4.

Remember to be cautious when multiplying or dividing by negative numbers or when working with absolute value inequalities, as these operations can change the direction of the inequality. Always check the sign of the coefficient and use appropriate properties to manipulate inequalities accurately.

The addition property for inequalities states that you can add (or subtract) the same value from both sides of an inequality without changing the direction of the inequality. Here are examples to illustrate this property:

Addition Property:

Example 1: Consider the inequality $3�-5<10$. To solve for $�$, you can add 5 to both sides:

$3�-5+5<10+5$

$3�<15$

Now, let's isolate $�$ by dividing both sides by 3 (a positive number):

$\frac{3�}{3}<\frac{15}{3}$

$�<5$

In this example, adding 5 to both sides of the inequality did not change the direction of the inequality. The original inequality $3�-5<10$ remains a "less than" inequality.

Example 2: Consider the inequality $-2�+7>3$. To solve for $�$, you can subtract 7 from both sides:

$-2�+7-7>3-7$

$-2�>-4$

Now, let's isolate $�$ by dividing both sides by -2 (a negative number):

$\frac{-2�}{-2}<\frac{-4}{-2}$

$�<2$

In this example, subtracting 7 from both sides of the inequality also did not change the direction of the inequality. The original inequality $-2�+7>3$ remains a "greater than" inequality.

The addition property for inequalities allows you to perform the same operation on both sides of the inequality, which is a useful tool for solving and manipulating inequalities.

The multiplication property for inequalities states that you can multiply both sides of an inequality by a positive number without changing the direction of the inequality. However, if you multiply both sides by a negative number, the direction of the inequality is reversed. Here are examples to illustrate this property:

Multiplying by a Positive Number:

Example 1: Suppose you have the inequality $2�<6$. To solve for $�$, you can divide both sides by 2 (which is positive):

$\frac{2�}{2}<\frac{6}{2}$

$�<3$

The direction of the inequality remains "less than" because we multiplied both sides by a positive number.

Multiplying by a Negative Number:

Example 2: Consider the inequality $-4�>8$. To solve for $�$, you can divide both sides by -4 (which is negative):

$\frac{-4�}{-4}>\frac{8}{-4}$

$�<-2$

In this case, the direction of the inequality was reversed from "greater than" to "less than" because we multiplied both sides by a negative number (-4).

It's important to remember that when multiplying or dividing by a negative number, you should reverse the direction of the inequality. When dealing with positive numbers, the direction remains unchanged. These principles are crucial for correctly solving and manipulating inequalities.

Solving inequalities in one variable algebraically involves finding the values of the variable that satisfy the inequality. Here are the steps to solve inequalities algebraically:

Step 1: Isolate the Variable:

Start by isolating the variable on one side of the inequality. This is similar to solving equations. Use addition, subtraction, multiplication, and division to achieve this.

Step 2: Consider Sign Changes:

Pay attention to sign changes when multiplying or dividing both sides by a negative number. If you multiply or divide by a negative number, reverse the direction of the inequality.

Step 3: Express the Solution:

Express the solution in interval notation or set notation, depending on the form of the solution.

Here are examples illustrating these steps:

Example 1: Solve the inequality $3�+2>8$.

Step 1: Isolate the variable $�$.

$3�+2>8$

Subtract 2 from both sides:

$3�>6$

Step 2: Consider sign changes (none needed in this case).

Step 3: Express the solution.

Divide both sides by 3:

$\frac{3�}{3}>\frac{6}{3}$

$�>2$

So, the solution is $�>2$, and it can be expressed in interval notation as $(2,\mathrm{\infty })$ or in set notation as $\{�\mid �>2\}$.

Example 2: Solve the inequality $-2(4-�)\le 6�-8$.

Step 1: Isolate the variable $�$.

$-2(4-�)\le 6�-8$

Distribute the -2 on the left side:

$-8+2�\le 6�-8$

Subtract 2x from both sides:

$-8\le 4�-8$

Step 2: Consider sign changes (none needed in this case).

Step 3: Express the solution.

Add 8 to both sides:

$-8+8\le 4�-8+8$

$0\le 4�$

Divide both sides by 4:

$\frac{0}{4}\le \frac{4�}{4}$

$0\le �$

So, the solution is $�\ge 0$, and it can be expressed in interval notation as $[0,\mathrm{\infty })$ or in set notation as $\{�\mid �\ge 0\}$.

These examples demonstrate the steps to algebraically solve inequalities in one variable. Remember to consider sign changes when necessary, especially when multiplying or dividing by negative numbers.

Solving inequalities involving fractions is similar to solving inequalities with integers or variables. You need to isolate the variable on one side of the inequality while keeping in mind the rules for working with fractions. Here are a couple of examples to illustrate how to solve inequalities with fractions:

Example 1: Solve the inequality $\frac{2�}{3}-\frac{1}{2}>\frac{1}{6}$.

Step 1: Clear the fractions by finding a common denominator. In this case, the common denominator is 6.

$\frac{2�}{3}-\frac{1}{2}>\frac{1}{6}$

Multiply both sides of the inequality by 6 to eliminate fractions:

$6(\frac{2�}{3}-\frac{1}{2})>6\left(\frac{1}{6}\right)$

This simplifies to:

$\frac{4�}{1}-\frac{3}{1}>1$

Step 2: Combine fractions and constants on the left side:

$\frac{4�-3}{1}>1$

Step 3: Isolate the variable by adding 3 to both sides:

$\frac{4�-3}{1}+3>1+3$

$\frac{4�-3+3}{1}>4$

$\frac{4�}{1}>4$

Step 4: Simplify the equation:

$4�>4$

Step 5: Divide by 4 (a positive number):

$\frac{4�}{4}>\frac{4}{4}$

$�>1$

So, the solution to the inequality is $�>1$, which can be expressed in interval notation as $(1,\mathrm{\infty })$ or in set notation as $\{�\mid �>1\}$.

Example 2: Solve the inequality $\frac{�+3}{2}\le \frac{5}{4}$.

Step 1: Clear the fractions by finding a common denominator. In this case, the common denominator is 4.

$\frac{�+3}{2}\le \frac{5}{4}$

Multiply both sides of the inequality by 4 to eliminate fractions:

$4\left(\frac{�+3}{2}\right)\le 4\left(\frac{5}{4}\right)$

This simplifies to:

$\frac{4(�+3)}{2}\le 5$

Step 2: Simplify the equation:

$\frac{2(�+3)}{1}\le 5$

Step 3: Isolate the variable by subtracting 3 from both sides:

$\frac{2(�+3)}{1}-3\le 5-3$

$\frac{2(�+3)-3}{1}\le 2$

Step 4: Simplify the equation:

$\frac{2�+6-3}{1}\le 2$

$\frac{2�+3}{1}\le 2$

Step 5: Divide by 2 (a positive number):

$\frac{2�+3}{2}\le \frac{2}{2}$

$�+\frac{3}{2}\le 1$

Step 6: Isolate $�$ by subtracting $\frac{3}{2}$ from both sides:

$�+\frac{3}{2}-\frac{3}{2}\le 1-\frac{3}{2}$

$�\le -\frac{1}{2}$

So, the solution to the inequality is $�\le -\frac{1}{2}$, which can be expressed in interval notation as $-\mathrm{\infty },-\frac{1}{2}]$ or in set notation as $\{�\mid �\le -\frac{1}{2}\}$.

These examples demonstrate how to solve inequalities with fractions by finding a common denominator and performing the same operations on both sides of the inequality to isolate the variable.

Compound inequalities are inequalities that involve more than one inequality statement connected by the words "and" or "or." They combine multiple conditions or inequalities into a single statement. There are two main types of compound inequalities: "and" inequalities and "or" inequalities.

1. "And" Inequalities:

In "and" inequalities, both conditions must be satisfied for the compound inequality to be true. They are often expressed using the symbol "∩" (intersection) or simply "and."

Example 1: Solve the compound inequality $2�-3<5$ and $�+1>-2$.

Here, we have two separate inequalities:

a) $2�-3<5$ b) $�+1>-2$

To solve the compound inequality, we need to satisfy both conditions simultaneously. So, we solve each inequality separately and then find the intersection of their solution sets.

For $2�-3<5$:

Add 3 to both sides: $2�<8$ Divide by 2 (positive): $�<4$

For $�+1>-2$:

Subtract 1 from both sides: $�>-3$

Now, find the intersection of the solution sets: $(-3,4)$.

So, the solution to the compound inequality is $-3<�<4$.

2. "Or" Inequalities:

In "or" inequalities, at least one of the conditions must be satisfied for the compound inequality to be true. They are often expressed using the symbol "∪" (union) or simply "or."

Example 2: Solve the compound inequality $2�-3<5$ or $�+1>7$.

Here, we have two separate inequalities:

a) $2�-3<5$ b) $�+1>7$

To solve the compound inequality, we need to satisfy at least one of the conditions. So, we solve each inequality separately and then find the union of their solution sets.

For $2�-3<5$:

Add 3 to both sides: $2�<8$ Divide by 2 (positive): $�<4$

For $�+1>7$:

Subtract 1 from both sides: $�>6$

Now, find the union of the solution sets: $(-\mathrm{\infty },4)\cup (6,\mathrm{\infty })$.

So, the solution to the compound inequality is $(-\mathrm{\infty },4)\cup (6,\mathrm{\infty })$.

Understanding and solving compound inequalities are essential skills in algebra and real-life applications, where multiple conditions need to be considered simultaneously or separately.

A compound inequality with the variable in all three parts typically involves an expression like "a < x < b," where $�$, $�$, and $�$ are real numbers. To solve such a compound inequality, you need to satisfy both inequalities simultaneously. Here's an example:

Example: Solve the compound inequality $1<2�-3<7$.

In this case, you have two inequalities combined by the "and" condition. You need to satisfy both conditions at the same time.

Step 1: Solve the first inequality, $1<2�-3$.

Add 3 to both sides:

$1+3<2�$

$4<2�$

Divide by 2 (a positive number, so the inequality sign remains the same):

$\frac{4}{2}<\frac{2�}{2}$

$2<�$

So, the solution to the first inequality is $2<�$.

Step 2: Solve the second inequality, $2�-3<7$.

Add 3 to both sides:

$2�-3+3<7+3$

$2�<10$

Divide by 2 (a positive number, so the inequality sign remains the same):

$\frac{2�}{2}<\frac{10}{2}$

$�<5$

So, the solution to the second inequality is $�<5$.

Step 3: Combine the solutions from Steps 1 and 2. Since both inequalities must be satisfied simultaneously, take the intersection of their solution sets:

$2<�$ (from Step 1)

$�<5$ (from Step 2)

Now, find the intersection of these intervals:

$2<�<5$

So, the solution to the compound inequality $1<2�-3<7$ is $2<�<5$.

This means that $�$ must be greater than 2 and less than 5 to satisfy the compound inequality.

Solving absolute value inequalities involves finding the values of the variable that satisfy an inequality containing absolute value expressions. These inequalities can have one of two forms: "less than" or "greater than" absolute value inequalities. Here's how to solve both types:

1. "Less Than" Absolute Value Inequalities:

The general form of a "less than" absolute value inequality is:

$\mathrm{\mid }�(�)\mathrm{\mid }<�$

where $�(�)$ is a function of $�$, and $�$ is a positive constant.

Steps to Solve "Less Than" Absolute Value Inequalities:

1. Set up two inequalities: one with the expression inside the absolute value bars positively and the other with the expression inside the absolute value bars negatively.

2. Solve both inequalities separately.

3. Combine the solutions from both inequalities.

Example: Solve the absolute value inequality $\mathrm{\mid }3�-2\mathrm{\mid }<4$.

a) $3�-2<4$

Add 2 to both sides:

$3�<6$

Divide by 3 (a positive number):

$�<2$

b) $-(3�-2)<4$

Distribute the negative sign:

$-3�+2<4$

Subtract 2 from both sides:

$-3�<2$

Divide by -3 (a negative number, so reverse the inequality sign):

$�>-\frac{2}{3}$

Combine the solutions: $-\frac{2}{3}<�<2$

So, the solution to $\mathrm{\mid }3�-2\mathrm{\mid }<4$ is $-\frac{2}{3}<�<2$.

2. "Greater Than" Absolute Value Inequalities:

The general form of a "greater than" absolute value inequality is:

$\mathrm{\mid }�(�)\mathrm{\mid }>�$

Steps to Solve "Greater Than" Absolute Value Inequalities:

1. Set up two inequalities: one with the expression inside the absolute value bars positively and the other with the expression inside the absolute value bars negatively.

2. Solve both inequalities separately.

3. Combine the solutions from both inequalities.

Example: Solve the absolute value inequality $\mathrm{\mid }2�+3\mathrm{\mid }>5$.

a) $2�+3>5$

Subtract 3 from both sides:

$2�>2$

Divide by 2 (a positive number):

$�>1$

b) $-(2�+3)>5$

Distribute the negative sign:

$-2�-3>5$

Add 3 to both sides:

$-2�>8$

Divide by -2 (a negative number, so reverse the inequality sign):

$�<-4$

Combine the solutions: $�<-4$ or $�>1$

So, the solution to $\mathrm{\mid }2�+3\mathrm{\mid }>5$ is $�<-4$ or $�>1$.

These examples illustrate how to solve both "less than" and "greater than" absolute value inequalities by considering both positive and negative cases for the absolute value expression and then combining the solutions.

Determining a number within a prescribed distance from another number involves finding all the values that are within a certain range or distance from a given number. This can be represented as an inequality.

Let's say you have a number $�$ and you want to find all numbers $�$ that are within a distance of $�$ units from $�$. The inequality to represent this situation is:

$\mathrm{\mid }�-�\mathrm{\mid }\le �$

This inequality states that the absolute value of the difference between $�$ and $�$ is less than or equal to $�$, meaning that $�$ is within $�$ units of $�$.

To solve this inequality, you'll consider two cases:

Case 1: $�-�\le �$

In this case, $�$ is to the right of $�$ on the number line and is within $�$ units of $�$. You will solve for $�$ by adding $�$ to both sides:

$�\le �+�$

Case 2: $-(�-�)\le �$

In this case, $�$ is to the left of $�$ on the number line and is within $�$ units of $�$. You will solve for $�$ by subtracting $�$ from both sides and changing the inequality sign:

$-�+�\le �$

To isolate $�$, multiply both sides by -1 (which reverses the inequality sign):

$�-�\ge -�$

Then, add $�$ to both sides:

$�\ge �-�$

So, the solution to the inequality $\mathrm{\mid }�-�\mathrm{\mid }\le �$ is:

$�-�\le �\le �+�$

This means that all values of $�$ within a distance of $�$ units from $�$ are between $�-�$ and $�+�$ on the number line.

For example, if you want to find all values of $�$ within 2 units of 5, the solution is $3\le �\le 7$, indicating that $�$ can be any number between 3 and 7 (including 3 and 7) to be within 2 units of 5.

A graphical approach to solving absolute value inequalities involves visualizing the solution set on a number line or a graph. Here are some examples illustrating this approach:

Example 1: Solve the absolute value inequality $\mathrm{\mid }�-2\mathrm{\mid }\le 3$.

Graphical Approach:

1. Start by drawing a number line.

2. Locate the point $�=2$ because $\mathrm{\mid }�-2\mathrm{\mid }$ means the absolute value of the difference between $�$ and 2.

3. Mark $�$ on the number line.

4. Next, consider the inequality $\mathrm{\mid }�-2\mathrm{\mid }\le 3$.

   a) For $�$ to be within 3 units of $�$, you need to mark points 3 units to the right and 3 units to the left of $�$.

   b) Mark points $2+3=5$ to the right of 2 and $2-3=-1$ to the left of 2.

   c) Now you have three points marked on the number line: -1, 2, and 5.

5. The solution to the inequality $\mathrm{\mid }�-2\mathrm{\mid }\le 3$ is all values of $�$ that fall within the interval between -1 and 5, including -1 and 5.

   
   

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Example 2: Solve the absolute value inequality $\mathrm{\mid }3�+1\mathrm{\mid }>4$.

Graphical Approach:

1. Draw a number line.

2. Locate the point $�=-\frac{1}{3}$ because $\mathrm{\mid }3�+1\mathrm{\mid }$ represents the absolute value of the expression $3�+1$.

3. Mark $�$ on the number line.

4. Consider the inequality $\mathrm{\mid }3�+1\mathrm{\mid }>4$.

   a) For $�$ to be more than 4 units away from $�$, mark points 4 units to the right and 4 units to the left of $�$.

   b) Mark points $-\frac{1}{3}+4=\frac{11}{3}$ to the right of $-\frac{1}{3}$ and $-\frac{1}{3}-4=-\frac{13}{3}$ to the left of $-\frac{1}{3}$.

   c) Now you have three points marked on the number line: $-\frac{13}{3}$, $-\frac{1}{3}$, and $\frac{11}{3}$.

5. The solution to the inequality $\mathrm{\mid }3�+1\mathrm{\mid }>4$ is all values of $�$ that fall outside the interval between $-\frac{13}{3}$ and $\frac{11}{3}$.

   
   

   $$\overline{)\begin{array}{ccc}\text{<span style="background-color: white;">Interval</span>}& \text{<span style="background-color: white;">Graphical Representation</span>}& \text{<span style="background-color: white;">Interval Notation</span>}\\ <span\; style="background-color:\; white;">(</span><span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">\infty </span>}<span\; style="background-color:\; white;">,</span><span\; style="background-color:\; white;">-</span>\frac{\mathrm{<span\; style="background-color:\; white;">13</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">\cup </span><span\; style="background-color:\; white;">(</span>\frac{\mathrm{<span\; style="background-color:\; white;">11</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">\infty </span>}<span\; style="background-color:\; white;">)</span>& \begin{array}{c}<span\; style="background-color:\; white;">(</span><span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">\infty </span>}<span\; style="background-color:\; white;">\circ </span><span\; style="background-color:\; white;">\circ </span><span\; style="background-color:\; white;">-</span>\frac{\mathrm{<span\; style="background-color:\; white;">13</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">\circ </span><span\; style="background-color:\; white;">\circ </span><span\; style="background-color:\; white;">-</span>\frac{\mathrm{<span\; style="background-color:\; white;">1</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">\circ </span><span\; style="background-color:\; white;">\circ </span>\frac{\mathrm{<span\; style="background-color:\; white;">11</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">\circ </span><span\; style="background-color:\; white;">\circ </span>\mathrm{<span\; style="background-color:\; white;">\infty </span>}<span\; style="background-color:\; white;">)</span>\end{array}& <span\; style="background-color:\; white;">(</span><span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">\infty </span>}<span\; style="background-color:\; white;">,</span><span\; style="background-color:\; white;">-</span>\frac{\mathrm{<span\; style="background-color:\; white;">13</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">\cup </span><span\; style="background-color:\; white;">(</span>\frac{\mathrm{<span\; style="background-color:\; white;">11</span>}}{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">\infty </span>}<span\; style="background-color:\; white;">)</span>\end{array}}$$

6. Chapter 2 Test:

7. 1. Graph the following: 4y=1x+2.

8. To graph the equation $4�=�+2$, we'll first need to rearrange it into slope-intercept form ($�=��+�$), where $�$ is the slope and $�$ is the y-intercept.

   Here's the equation in slope-intercept form:

   $4�=�+2$

   Divide both sides by 4 to isolate $�$:

   $�=\frac{1}{4}�+\frac{2}{4}$

   Simplify:

   $�=\frac{1}{4}�+\frac{1}{2}$

   Now, you can see that the equation is in slope-intercept form, where the slope ($�$) is $\frac{1}{4}$ and the y-intercept ($�$) is $\frac{1}{2}$.

   To graph it, follow these steps:

   1. Plot the y-intercept: The y-intercept is the point (0, 1/2). So, start by marking this point on the coordinate plane.

   2. Use the slope: The slope $\frac{1}{4}$ means that for every 1 unit you move to the right (positive x-direction), you should move $\frac{1}{4}$ units up. Starting from the y-intercept, move 1 unit to the right and $\frac{1}{4}$ unit up. Plot this point.

   3. Repeat the process: Keep moving 1 unit to the right and $\frac{1}{4}$ unit up from the last point. Plot several more points in this manner.

   4. Connect the points: Once you've plotted enough points, draw a straight line through them. This line represents the graph of the equation $4�=�+2$.

   Your graph should look like a line sloping upward from left to right.

9. 2. Find the x- and y-intercepts for the following: 1x−4y=7

10. To find the x- and y-intercepts of the equation $�-4�=7$, you can set either $�$ or $�$ to 0 and solve for the other variable.

    X-Intercept:

    To find the x-intercept, set $�$ to 0 and solve for $�$:

    $�-4(0)=7$

    Simplify:

    $�=7$

    So, the x-intercept is 7, which can be written as the point (7, 0).

    Y-Intercept:

    To find the y-intercept, set $�$ to 0 and solve for $�$:

    $0-4�=7$

    Simplify:

    $-4�=7$

    Divide both sides by -4 to isolate $�$:

    $�=-\frac{7}{4}$

    So, the y-intercept is $-\frac{7}{4}$, which can be written as the point (0, -7/4).

    Therefore, the x-intercept is (7, 0), and the y-intercept is (0, -7/4) for the equation $�-4�=7$.

11. 3. Find the x- and y-intercepts of this equation, and sketch the graph of the line using just the intercepts plotted. 3x−4y=14

12. 
    

13. To find the x- and y-intercepts of the equation $3�-4�=14$, you can set either $�$ or $�$ to 0 and solve for the other variable.

    X-Intercept:

    To find the x-intercept, set $�$ to 0 and solve for $�$:

    $3�-4(0)=14$

    Simplify:

    $3�=14$

    Divide both sides by 3 to isolate $�$:

    $�=\frac{14}{3}$

    So, the x-intercept is $\frac{14}{3}$, which can be written as the point $(\frac{14}{3},0)$.

    Y-Intercept:

    To find the y-intercept, set $�$ to 0 and solve for $�$:

    $3(0)-4�=14$

    Simplify:

    $-4�=14$

    Divide both sides by -4 to isolate $�$:

    $�=-\frac{14}{4}$

    Simplify:

    $�=-\frac{7}{2}$

    So, the y-intercept is $-\frac{7}{2}$, which can be written as the point $(0,-\frac{7}{2})$.

    Now, let's sketch the graph using just the intercepts plotted:

    Plot the x-intercept $(\frac{14}{3},0)$ and the y-intercept $(0,-\frac{7}{2})$ on a coordinate plane.

    Then, draw a straight line through these two points. This line represents the graph of the equation $3�-4�=14$.

    The graph should look like a line sloping upward from left to right, passing through the points $(\frac{14}{3},0)$ and $(0,-\frac{7}{2})$.

4. Find the exact distance between  (5,−3)

  and  (−2,8).

  Find the coordinates of the midpoint of the line segment joining the two points.

To find the exact distance between two points $({�}_1,{�}_1)$ and $({�}_2,{�}_2)$, you can use the distance formula:

$�=\sqrt{({�}_2-{�}_1{)}^2+({�}_2-{�}_1{)}^2}$

In your case, the two points are $(5,-3)$ and $(-2,8)$, so ${�}_1=5$, ${�}_1=-3$, ${�}_2=-2$, and ${�}_2=8$. Plug these values into the distance formula:

$�=\sqrt{(-2-5{)}^2+(8-(-3){)}^2}$

$�=\sqrt{(-7{)}^2+(8+3{)}^2}$

$�=\sqrt{49+121}$

$�=\sqrt{170}$

Now, let's simplify $\sqrt{170}$. Since 170 is not a perfect square, we leave it in simplified radical form:

$�=\sqrt{170}$

So, the exact distance between the points $(5,-3)$ and $(-2,8)$ is $\sqrt{170}$.

To find the coordinates of the midpoint of the line segment joining these two points, you can use the midpoint formula:

$({�}_{�},{�}_{�})=(\frac{{�}_1+{�}_2}{2},\frac{{�}_1+{�}_2}{2})$

In this case:

$({�}_{�},{�}_{�})=(\frac{5+(-2)}{2},\frac{(-3)+8}{2})$

Now, calculate the coordinates of the midpoint:

$({�}_{�},{�}_{�})=(\frac{3}{2},\frac{5}{2})$

So, the coordinates of the midpoint of the line segment joining the points $(5,-3)$ and $(-2,8)$ are $(\frac{3}{2},\frac{5}{2})$.

5. Write the interval notation for the set of numbers represented by {x|x≤9}.

The interval notation for the set of numbers represented by $\{�\mathrm{\mid }�\le 9\}$ is:

$-\mathrm{\infty }<�\le 9$

This notation indicates that $�$ can be any real number greater than negative infinity but less than or equal to 9.

6. Solve for x: 5x+7=3x−11.

To solve for $�$ in the equation $5�+7=3�-11$, follow these steps:

Step 1: Subtract $3�$ from both sides of the equation to isolate the $�$ terms on one side.

$5�-3�+7=3�-3�-11$

Simplify:

$2�+7=-11$

Step 2: Subtract 7 from both sides to isolate the $2�$ term.

$2�+7-7=-11-7$

Simplify:

$2�=-18$

Step 3: Finally, divide both sides by 2 to solve for $�$.

$\frac{2�}{2}=\frac{-18}{2}$

Simplify:

$�=-9$

So, the solution to the equation is $�=-9$.

7. Solve for x : 3(2x−5)−3(x−7)=2x−9.

To solve for $�$ in the equation $3(2�-5)-3(�-7)=2�-9$, follow these steps:

Step 1: Distribute the constants inside the parentheses:

$6�-15-3�+21=2�-9$

Step 2: Combine like terms on each side of the equation:

On the left side, combine $6�$ and $-3�$:

$6�-3�=3�$

On the left side, combine $-15$ and $21$:

$-15+21=6$

So, the equation simplifies to:

$3�+6=2�-9$

Step 3: Subtract $2�$ from both sides to isolate $�$ terms on the right side:

$3�-2�+6=-9$

Simplify:

$�+6=-9$

Step 4: Subtract 6 from both sides to isolate $�$:

$�+6-6=-9-6$

Simplify:

$�=-15$

So, the solution to the equation is $�=-15$.

8. Solve for x: x2+1=4x

To solve for $�$ in the equation ${�}^2+1=4�$, follow these steps:

Step 1: Move all the terms to one side of the equation to set it equal to zero:

${�}^2-4�+1=0$

Step 2: Try to factor the quadratic expression. In this case, factoring may not be straightforward, so we can use the quadratic formula:

The quadratic formula is: $�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

In this equation, $�=1$, $�=-4$, and $�=1$.

Step 3: Plug the values into the quadratic formula:

$�=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1)(1)}}{2(1)}$

Simplify:

$�=\frac{4\pm \sqrt{16-4}}{2}$

$�=\frac{4\pm \sqrt{12}}{2}$

Step 4: Simplify further:

$�=\frac{4\pm 2\sqrt{3}}{2}$

Step 5: Factor out a common factor of 2 in the numerator:

$�=\frac{2(2\pm \sqrt{3})}{2}$

Step 6: Cancel the common factor of 2:

$�=2\pm \sqrt{3}$

So, the solutions to the equation ${�}^2+1=4�$ are $�=2+\sqrt{3}$ and $�=2-\sqrt{3}$.