Solving For X In The Equation 4(px + 1) = 64

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This article provides a step-by-step solution for finding the value of x in the equation 4(px + 1) = 64, both in terms of p and when p has a specific value. This is a fundamental algebraic problem that involves distributing, simplifying, and isolating the variable x. Understanding how to solve such equations is crucial for various mathematical and scientific applications. Let's dive into the process of solving this equation methodically.

Finding x in terms of p

To determine the value of x in terms of p, we need to isolate x on one side of the equation. The initial equation is:

4(px + 1) = 64

The first step is to distribute the 4 across the terms inside the parentheses:

4px + 4 = 64

Next, we subtract 4 from both sides of the equation to isolate the term containing x:

4px = 64 - 4 4px = 60

Now, we divide both sides by 4p to solve for x:

x = 60 / (4p)

Simplify the fraction by dividing both the numerator and the denominator by 4:

x = 15 / p

Thus, the value of x in terms of p is x = 15/p. This expression tells us that the value of x depends inversely on the value of p. As p increases, x decreases, and vice versa. This relationship is essential to understand when dealing with such algebraic expressions.

Detailed Explanation of Each Step

  1. Distribution: The initial step involves distributing the constant 4 across the terms inside the parentheses (px and 1). This is based on the distributive property of multiplication over addition, which states that a(b + c) = ab + ac. In our case, 4(px + 1) becomes 4px + 4.
  2. Isolating the Term with x: To isolate the term containing x (which is 4px), we subtract 4 from both sides of the equation. This maintains the balance of the equation while moving the constant term to the right side. The equation 4px + 4 = 64 becomes 4px = 60.
  3. Solving for x: To finally solve for x, we divide both sides of the equation by 4p. This isolates x on the left side, giving us x = 60 / (4p). It is crucial to divide by the entire term 4p to correctly isolate x.
  4. Simplification: The final step involves simplifying the fraction 60 / (4p). We can divide both the numerator and the denominator by their greatest common factor, which is 4. This simplifies the fraction to 15 / p, giving us the final expression for x in terms of p: x = 15/p.

This detailed breakdown of each step helps to understand the logical progression and the algebraic principles applied in solving for x in terms of p. By following these steps, you can confidently solve similar equations and manipulate algebraic expressions.

Finding the Value of x when p = -5

Now that we have x in terms of p, we can easily find the value of x when p = -5. We substitute -5 for p in the expression we derived:

x = 15 / p x = 15 / (-5)

Dividing 15 by -5, we get:

x = -3

Therefore, when p = -5, the value of x is -3. This demonstrates how the expression x = 15/p allows us to find x for any given value of p, making it a versatile solution to the original equation.

Implications of p = -5 on the Equation

When p = -5, the equation 4(px + 1) = 64 becomes 4(-5x + 1) = 64. Substituting -5 for p gives us a specific instance of the original equation, which we can then solve for x. This is a common practice in algebra, where we substitute known values into an equation to find the value of the unknown variable.

The process of substituting p = -5 and solving for x involves the same algebraic principles we used earlier:

  1. Substitution: We replace p with -5 in the equation, resulting in 4(-5x + 1) = 64.
  2. Distribution: Distribute the 4 across the terms inside the parentheses: 4(-5x) + 4(1) = -20x + 4.
  3. Equation Setup: The equation now looks like -20x + 4 = 64.
  4. Isolating the Term with x: Subtract 4 from both sides to isolate the term containing x: -20x = 60.
  5. Solving for x: Divide both sides by -20 to solve for x: x = 60 / -20.
  6. Simplification: Simplify the fraction: x = -3.

This step-by-step approach ensures that we correctly apply the algebraic principles and arrive at the correct value of x. The result, x = -3, is the solution to the original equation when p is specifically -5.

Importance of Understanding Variable Relationships

Understanding the relationship between variables in an equation is a fundamental concept in algebra. In this case, we explored how x is related to p in the equation 4(px + 1) = 64. By expressing x in terms of p, we gained insight into how changes in p affect the value of x. This understanding is crucial for solving more complex equations and problems in mathematics and other fields.

The relationship x = 15/p shows an inverse proportionality between x and p. This means that as p increases, x decreases, and vice versa. This type of relationship is common in various scientific and engineering contexts. For example, in physics, the relationship between pressure and volume of a gas at constant temperature follows a similar inverse proportionality (Boyle's Law).

Real-World Applications of Variable Relationships

Understanding variable relationships is not just a theoretical exercise; it has practical applications in many real-world scenarios:

  1. Physics: In physics, understanding the relationship between force, mass, and acceleration (Newton's Second Law) is crucial for analyzing the motion of objects.
  2. Economics: In economics, understanding the relationship between supply, demand, and price is essential for making informed decisions in the market.
  3. Engineering: In engineering, understanding the relationship between stress, strain, and material properties is vital for designing safe and efficient structures.
  4. Computer Science: In computer science, understanding the relationship between algorithm complexity and input size is necessary for optimizing software performance.

By mastering the concept of variable relationships, you can better understand and analyze the world around you. Whether you are solving a mathematical problem, designing a bridge, or developing a software application, the ability to see how variables interact is a valuable skill.

In summary, we have successfully found the value of x in terms of p to be x = 15/p, and when p = -5, the value of x is -3. This exercise demonstrates the importance of algebraic manipulation and the ability to solve equations for unknown variables. By understanding the steps involved in isolating variables and substituting values, you can tackle a wide range of algebraic problems. Remember, the key is to follow a systematic approach, applying algebraic principles correctly and simplifying expressions whenever possible. This foundation will serve you well in more advanced mathematical studies and practical applications.