Problem 14 - College Algebra

Question

Solve the inequality. Write your final answer in interval notation. $$ \frac{x-1}{3}+\frac{x+2}{5} \leq \frac{3}{5} $$

Step-by-Step Solution

Answer

The solution is \((-\infty, 1]\).

Step 1: Combine Fractions with a Common Denominator

To combine the fractions, first find a common denominator, which in this case is 15. Rewriting each fraction:\[\frac{x-1}{3} = \frac{5(x-1)}{15}, \frac{x+2}{5} = \frac{3(x+2)}{15}, \frac{3}{5} = \frac{9}{15}\] This gives us the inequality:\[\frac{5(x-1) + 3(x+2)}{15} \leq \frac{9}{15}.\]

Step 2: Clear the Fractions

Multiply both sides of the inequality by 15 to eliminate the denominators:\[5(x-1) + 3(x+2) \leq 9.\] This removes the fraction and simplifies the inequality.

Step 3: Distribute and Combine Like Terms

Distribute the constants and combine like terms:\[5x - 5 + 3x + 6 \leq 9.\]Combine like terms:\[8x + 1 \leq 9.\]

Step 4: Isolate the Variable

Subtract 1 from both sides to isolate the terms with \(x\):\[8x \leq 8.\]Then divide each side by 8 to solve for \(x\):\[x \leq 1.\]

Step 5: Write the Solution in Interval Notation

The solution to the inequality is all \(x\) values less than or equal to 1. In interval notation, this is:\[(-\infty, 1].\]

Key Concepts

Interval Notation

Interval notation is a way of writing sets of numbers to describe solutions in a condensed form. It is particularly useful in expressing the solution of inequalities. When dealing with inequalities like the one in our exercise, interval notation denotes all possible values for the variable that satisfy the inequality.
For instance, if an inequality solution includes all numbers less than or equal to 1, we write this in interval notation as \((-\infty, 1] \).

• The round parenthesis \(( \) denotes that the endpoint is not included, often used with infinity.
• The square bracket \([ \) indicates that the endpoint is included.

Thus, we use round brackets with infinity symbols, as infinity is not a specific number, whereas real numbers that are included use square brackets.

Rational Inequalities

Rational inequalities are inequalities that feature rational expressions, such as fractions where polynomials appear in the numerator and/or the denominator. Solving these involves handling each fraction by finding common denominators and combining them effectively.
In our exercise, the inequality \(\frac{x-1}{3}+\frac{x+2}{5} \leq \frac{3}{5}\) is an example of a rational inequality. Here, the goal is to simplify these expressions, ultimately allowing for easier manipulation and solution.
Key steps include:

• Finding a common denominator for the fractions involved, which allows you to perform operations like addition or comparison.
• Eliminating the fractions to handle simpler, polynomial inequalities.

Understanding this process makes solving rational inequalities systematic and manageable.

Combining Fractions

Combining fractions involves rewriting fractions so that they have a common denominator. This allows them to be added or subtracted. Achieving this is a crucial step in solving rational inequalities as shown in our example.
To combine fractions:

• Identify the least common denominator (LCD). For example, between 3 and 5, the LCD is 15.
• Reexpress each fraction with the LCD. For instance, \(\frac{x-1}{3} \) becomes \(\frac{5(x-1)}{15}\).
• Perform the operations, such as addition, once the fractions have the same denominators.

This technique simplifies complex expressions and is essential for solving more involved algebraic problems involving fractions.

Variable Isolation

Variable isolation is separating the variable of interest from other numbers or coefficients in an equation or inequality. This is typically done to solve for the variable's values.
In our specific inequality, starting with the equation \(8x + 1 \leq 9\), we want to get \(x\) alone.
We follow these steps:

• First, subtract 1 from both sides to begin isolating \(x\). This changes the inequality to \(8x \leq 8\).
• Next, divide each side by 8, resulting in \(x \leq 1\).

Using these straightforward operations, variable isolation helps simplify equations and allows one to find solutions with ease.