Problem 7 - College Algebra

Question

Describe the situation in which the distance that point \(x\) is from 10 is at least 15 units. Express this using absolute value notation.

Step-by-Step Solution

Answer

The situation is expressed as \(|x - 10| \geq 15\).

Step 1: Understanding the Problem

We need to find situations where the distance from the point, represented by variable \(x\), to the number 10 is at least 15 units. This situation is commonly expressed using absolute value notation, which represents the distance from a number on the real number line.

Step 2: Setting up the Absolute Value Equation

The absolute value notation for the distance from \(x\) to 10 being at least 15 units is \(|x - 10| \geq 15\). This notation means that whether \(x\) is to the left or to the right of 10 on the number line, the distance is 15 units or more.

Step 3: Interpreting the Equation

The equation \(|x - 10| \geq 15\) can be split into two separate inequalities without the absolute value: \(x - 10 \geq 15\) and \(x - 10 \leq -15\). This tells us that either \(x\) is 15 units or more to the right of 10, or \(x\) is 15 units or more to the left of 10.

Step 4: Solving the First Inequality

Starting with the inequality \(x - 10 \geq 15\), solve for \(x\) by adding 10 to both sides: \(x \geq 25\). This means that one possible situation is \(x\) being 25 or greater.

Step 5: Solving the Second Inequality

For the inequality \(x - 10 \leq -15\), solve for \(x\) by adding 10 to both sides: \(x \leq -5\). This means another possible situation is \(x\) being -5 or less.

Step 6: Combining the Results

The solutions from both inequalities indicate that the point \(x\) must either be greater than or equal to 25 or less than or equal to -5 for the distance from 10 to be at least 15 units.

Key Concepts

Distance on Number Line

When dealing with absolute value equations, it's often helpful to visualize the problem on a number line. The number line is a straight line where each point represents a real number, allowing us to see distances between numbers in an easy and clear manner.

In the context of the problem, the absolute value \(|x - 10|\) represents the distance from some point \(x\) to the number 10 on this number line.

Distance, in mathematics, is always a non-negative value, meaning it doesn't matter if \(x\) is to the left or right of 10; we only care about how far apart they are.

• If \(x\) lies exactly 15 units to the right of 10, then \(x = 25\).
• If \(x\) is 15 units to the left of 10, then \(x = -5\).

Thus, when we say the distance is "at least 15 units," it implies that \(x\) could also be further than \(-5\) or \(25\), covering all numbers beyond these points.

Inequalities

Inequalities are mathematical expressions that show the relationship between two values where they are not necessarily equal. They are crucial for interpreting absolute value equations involving distance.

In our exercise, after expressing the absolute value equation \(|x - 10| \geq 15\), we translate it to two separate inequalities to find possible values for \(x\):

• \(x - 10 \geq 15\) leads to \(x \geq 25\).
• \(x - 10 \leq -15\) results in \(x \leq -5\).

These inequalities help depict where the point \(x\) can lie on the number line to satisfy the original absolute value condition.

The solution space indicated by these inequalities tells us about possible (and sometimes infinite) solutions that fulfill the given condition.

Algebraic Expressions

Algebraic expressions are collections of numbers, variables, and operations (like addition or subtraction) that represent a value. In the exercise, our key expression \(|x - 10|\) involves subtraction, with \(x\) being a variable and 10 a constant.

Breaking down the absolute value expression, \(x - 10\) is the algebraic part prior to applying the absolute value.

• It directly compares \(x\) to 10, which is crucial for the distance interpretation.
• The subtraction indicates moving right or left from 10 on the number line.

Transforming the absolute expression to inequalities, \(x - 10 \geq 15\) and \(x - 10 \leq -15\), allows us to solve algebraically by performing arithmetic operations.

These operations essentially help in expressing ranges of numbers that \(x\) can be, adhering to the distance requirement stated in the problem.