Solving Inequalities A Step-by-Step Guide To 6 + Y/3 ≤ 15

In this article, we will delve into the process of solving the inequality 6 + y/3 ≤ 15. Inequalities are mathematical statements that compare two expressions using symbols like less than ( < ), greater than ( > ), less than or equal to ( ≤ ), and greater than or equal to ( ≥ ). Solving an inequality involves finding the range of values for the variable that make the inequality true. This particular inequality involves a variable, y, a fraction, and a couple of constants, making it a good example for illustrating the steps involved in solving linear inequalities. We will walk through each step meticulously, ensuring a clear understanding of the underlying principles. This exploration will not only help in solving this specific problem but also lay a strong foundation for tackling more complex inequalities in the future.

Before we dive into solving the specific inequality 6 + y/3 ≤ 15, it's crucial to grasp the fundamental concept of inequalities. Unlike equations, which assert the equality of two expressions, inequalities express a relationship where one expression is either greater than, less than, greater than or equal to, or less than or equal to another. This distinction leads to a range of solutions rather than a single value, which is common in equations. Understanding this difference is key to approaching and solving inequalities effectively. When we encounter an inequality, we are essentially looking for a set of numbers that, when substituted for the variable, will satisfy the given condition. For example, in the inequality x > 5, any number greater than 5 will make the statement true. This set of numbers is known as the solution set. The solution set can be represented graphically on a number line, providing a visual representation of all possible values that satisfy the inequality. In this article, we'll apply these foundational concepts to the inequality 6 + y/3 ≤ 15, demonstrating how to isolate the variable and determine the range of values that fulfill the inequality.

To solve the inequality 6 + y/3 ≤ 15, we will follow a series of algebraic steps to isolate the variable y. This process is similar to solving equations, but with a crucial difference to keep in mind when multiplying or dividing by a negative number. Our goal is to get y by itself on one side of the inequality, revealing the range of values that satisfy the condition. Let's break down the solution step-by-step:

Step 1: Isolate the term with the variable

The first step in solving the inequality 6 + y/3 ≤ 15 is to isolate the term containing the variable, which in this case is y/3. To do this, we need to eliminate the constant term, 6, from the left side of the inequality. We achieve this by subtracting 6 from both sides of the inequality. This operation maintains the balance of the inequality, just like it does in an equation. By subtracting 6 from both sides, we effectively move the constant term to the right side, bringing us closer to isolating y. This step is crucial because it simplifies the inequality, making it easier to work with and ultimately solve for the variable. The result of this subtraction will give us a new, simplified inequality that we can then manipulate further to find the solution. Remember, the key is to perform the same operation on both sides to maintain the integrity of the inequality.

6 + y/3 ≤ 15
6 + y/3 - 6 ≤ 15 - 6
y/3 ≤ 9

Step 2: Multiply both sides by 3

Now that we have isolated the term y/3, the next step is to isolate y itself. Since y is being divided by 3, we need to perform the inverse operation, which is multiplication. We multiply both sides of the inequality by 3. This will cancel out the division by 3 on the left side, leaving us with just y. It's important to remember that when we multiply or divide both sides of an inequality by a positive number, the direction of the inequality sign remains the same. This is a fundamental rule in solving inequalities. By multiplying both sides by 3, we are essentially scaling up both sides of the inequality equally, which preserves the relationship between them. This step is critical in our quest to find the value or range of values for y that satisfy the original inequality. After this multiplication, we will have y isolated on one side, and a numerical value on the other, giving us the solution to the inequality.

y/3 ≤ 9
(y/3) * 3 ≤ 9 * 3
y ≤ 27

After performing the necessary algebraic steps, we have arrived at the solution to the inequality 6 + y/3 ≤ 15. The solution is y ≤ 27. This means that any value of y that is less than or equal to 27 will satisfy the original inequality. It's important to understand that this solution is not a single value, but rather a range of values. This is a key characteristic of inequalities, distinguishing them from equations which typically have a single solution. The solution y ≤ 27 represents an infinite number of values, all of which make the inequality true. For instance, if we substitute y = 27 into the original inequality, we get 6 + 27/3 ≤ 15, which simplifies to 6 + 9 ≤ 15, or 15 ≤ 15, a true statement. Similarly, any value less than 27, such as 0, 10, or 20, will also satisfy the inequality. This range of solutions is a fundamental aspect of working with inequalities, and understanding this concept is crucial for solving more complex problems.

To ensure the accuracy of our solution, it is always a good practice to verify it. We can verify the solution y ≤ 27 by substituting a value within the range and a value outside the range into the original inequality, 6 + y/3 ≤ 15. This process will help us confirm that our solution is correct and that we haven't made any errors in our algebraic manipulations.

Testing a value within the range (y = 21)

Let's first substitute a value within the solution range, say y = 21, into the original inequality. This will allow us to see if the inequality holds true for a value that our solution suggests should work. Substituting y = 21 into 6 + y/3 ≤ 15, we get 6 + 21/3 ≤ 15. Simplifying this, we have 6 + 7 ≤ 15, which further simplifies to 13 ≤ 15. This statement is true, as 13 is indeed less than or equal to 15. This verification step provides confidence that our solution is on the right track. However, to be completely sure, we also need to test a value outside the solution range.

Testing a value outside the range (y = 30)

Now, let's substitute a value outside the solution range, say y = 30, into the original inequality. This will help us confirm that values outside our solution do not satisfy the inequality, further validating our result. Substituting y = 30 into 6 + y/3 ≤ 15, we get 6 + 30/3 ≤ 15. Simplifying this, we have 6 + 10 ≤ 15, which further simplifies to 16 ≤ 15. This statement is false, as 16 is not less than or equal to 15. This confirms that values outside our solution range do not satisfy the inequality. By testing both a value within the range and a value outside the range, we have successfully verified that our solution y ≤ 27 is correct. This verification process is a crucial step in problem-solving, ensuring the accuracy of our results.

Visualizing the solution to an inequality can provide a deeper understanding of the range of values that satisfy it. The solution y ≤ 27 can be graphically represented on a number line. This representation helps to illustrate that the solution includes all numbers less than or equal to 27. To represent this on a number line, we draw a horizontal line and mark the number 27. Since the inequality includes