Problem 114 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve: \(\quad 1.3=P \cdot 26\)

Step-by-Step Solution

Answer

The solution to the equation \(1.3 = P \cdot 26\) is \(P \approx 0.05\).

Step 1: Understand the problem

The problem asks to solve for P in the equation \(1.3 = P \cdot 26\). To find the value of P, it is necessary to isolate P on one side of the equation.

Step 2: Divide by 26

To isolate P, divide both sides of the equation by 26. This means taking \(1.3 = P \cdot 26\) and divide both sides by 26 to get \(P = \frac{1.3}{26}\).

Step 3: Perform the division

Divide 1.3 by 26 to find the value of P. This results in P being approximately 0.05.

Key Concepts

Isolating Variables

In algebra, solving equations requires isolating the variable you are interested in, which means getting it alone on one side of the equation. This technique allows us to find the unknown value by transforming the equation step-by-step.

The first thing we need to do is identify the variable we want to isolate. In the exercise provided, the goal is to solve for \( P \). To do this, we aim to move everything else on the equation's side featuring \( P \) to the opposite side by performing inverse operations.

Let's consider the equation \( 1.3 = P \cdot 26 \). To isolate \( P \), we recognize that it is currently being multiplied by 26. Therefore, we perform the opposite operation, which is division, to both sides of the equation. By dividing both sides by 26, \( P \) becomes isolated, resulting in \( P = \frac{1.3}{26} \).

The process of isolation is crucial in algebra because it simplifies the problem to finding unknowns. By focusing only on one variable at a time, we better understand how different parts of the equation interact.

Division in Algebra

Division is a fundamental operation used in algebra to solve equations. It is especially useful when you need to isolate variables that are multiplied by numbers, as seen in the example equation \( 1.3 = P \cdot 26 \).

When dividing in algebra, remember the following important points:

• Division is the inverse of multiplication. It helps to "undo" multiplication and move terms around in the equation.
• Always perform the division operation on both sides of the equation. This maintains the equality, ensuring one side doesn't become larger or smaller than the other.
• Be precise. Division relies on accurate calculations to provide the right solution.

For our exercise, dividing both sides of \( 1.3 = P \cdot 26 \) by 26 gives us \( P = \frac{1.3}{26} \).

Dividing numbers can sometimes be tricky, especially with decimals and fractions. Practicing division regularly helps to build accuracy and confidence in solving algebraic equations. This method of "undoing" multiplication is integral to solving for variables effectively.

Finding Values of Unknowns

In algebra, the primary objective of equation solving is to find the value of unknown variables. These unknowns, often represented by letters such as \( P \), are the key to understanding relationships in mathematical expressions.

To find the value of an unknown, follow these steps:

• First, isolate the unknown variable in the equation, as previously discussed.
• Use appropriate mathematical operations like addition, subtraction, multiplication, or division to simplify the equation.
• Perform calculations accurately to determine the value of the unknown.

In our exercise example, the equation \( 1.3 = P \cdot 26 \) required us to isolate \( P \). By dividing both sides by 26, we simplified it to \( P = \frac{1.3}{26} \). Upon calculation, we found \( P \approx 0.05 \).

This process helps make abstract concepts tangible and answers practical questions. Finding unknowns is central to solving real-world problems, ensuring that we not only understand the equation but also its applications in various contexts.