Solving The Equation 5n/2 + 15 = 35 A Comprehensive Guide
In the realm of mathematics, mastering the art of solving linear equations is a fundamental skill. Linear equations, characterized by a variable raised to the power of one, form the bedrock of numerous mathematical concepts and applications. This article delves into the intricacies of solving the linear equation 5n/2 + 15 = 35, providing a comprehensive step-by-step guide that will empower you to tackle similar equations with confidence.
Understanding Linear Equations
Before we embark on the solution, let's first grasp the essence of linear equations. A linear equation is an algebraic equation where the highest power of the variable is one. These equations can be expressed in the general form ax + b = c, where 'a', 'b', and 'c' are constants, and 'x' is the variable we aim to solve for. The equation 5n/2 + 15 = 35 perfectly aligns with this form, making it a prime example of a linear equation.
Isolating the Variable Term: The First Crucial Step
The cornerstone of solving any linear equation lies in isolating the variable term. In our equation, 5n/2 + 15 = 35, the variable term is 5n/2. To isolate this term, we must eliminate the constant term, which is 15. This can be achieved by performing the inverse operation of addition, which is subtraction. Subtracting 15 from both sides of the equation maintains the balance and yields:
5n/2 + 15 - 15 = 35 - 15
Simplifying the equation, we arrive at:
5n/2 = 20
Eliminating the Fraction: A Necessary Transformation
The presence of a fraction, 5n/2, can initially seem daunting, but it's a hurdle easily overcome. To eliminate the fraction, we employ the inverse operation of division, which is multiplication. Multiplying both sides of the equation by the denominator, 2, ensures the fraction vanishes, paving the way for further simplification:
(5n/2) * 2 = 20 * 2
This simplifies to:
5n = 40
Solving for the Variable: The Final Revelation
With the variable term now isolated, the final step involves solving for 'n'. The variable 'n' is currently multiplied by 5. To isolate 'n', we invoke the inverse operation of multiplication, which is division. Dividing both sides of the equation by 5 unveils the solution:
5n / 5 = 40 / 5
Therefore, the solution to the equation is:
n = 8
Verifying the Solution: Ensuring Accuracy
The pursuit of accuracy is paramount in mathematics. To ensure our solution is correct, we substitute n = 8 back into the original equation, 5n/2 + 15 = 35, and verify if the equation holds true:
(5 * 8) / 2 + 15 = 35
Simplifying the equation:
40 / 2 + 15 = 35
20 + 15 = 35
35 = 35
The equation holds true, confirming that our solution, n = 8, is indeed correct.
Conclusion: Mastering the Art of Linear Equations
Solving linear equations is a fundamental skill that unlocks doors to a myriad of mathematical concepts and applications. By systematically isolating the variable term, eliminating fractions, and employing inverse operations, we can confidently navigate the world of linear equations. This comprehensive guide has equipped you with the knowledge and tools to solve equations like 5n/2 + 15 = 35 and beyond. Embrace the challenge, practice diligently, and watch your mathematical prowess flourish.
In our quest to master mathematical problem-solving, understanding the nuances of linear equations is paramount. Let's delve deeper into the equation 5n/2 + 15 = 35, dissecting each step and exploring the underlying principles that govern its solution. This in-depth analysis will not only solidify your understanding of this specific equation but also equip you with the skills to tackle a broader range of mathematical challenges.
Reaffirming the Foundation: The Essence of Linear Equations
As we embark on this deeper exploration, let's reaffirm the foundational concept of linear equations. A linear equation, in its simplest form, is an algebraic equation where the highest power of the variable is one. These equations can be represented in the general form ax + b = c, where 'a', 'b', and 'c' are constants, and 'x' is the variable we seek to determine. The equation 5n/2 + 15 = 35 aligns perfectly with this definition, making it an ideal candidate for our comprehensive analysis.
Isolating the Variable Term: A Strategic Maneuver
The cornerstone of solving any linear equation lies in isolating the variable term, the term containing the unknown variable. In our equation, 5n/2 + 15 = 35, the variable term is 5n/2. To isolate this term, we must strategically eliminate the constant term, which is 15. This is accomplished by employing the inverse operation of addition, which is subtraction. Subtracting 15 from both sides of the equation maintains the delicate balance and yields:
5n/2 + 15 - 15 = 35 - 15
Simplifying the equation, we arrive at:
5n/2 = 20
This step exemplifies the fundamental principle of maintaining equality in equations. Any operation performed on one side of the equation must be mirrored on the other side to preserve the balance and ensure the integrity of the solution.
Taming the Fraction: A Necessary Transformation
The presence of a fraction, 5n/2, can initially appear as an obstacle, but it's a hurdle easily overcome with the right approach. To eliminate the fraction, we leverage the inverse operation of division, which is multiplication. Multiplying both sides of the equation by the denominator, 2, effectively cancels out the fraction, paving the way for further simplification:
(5n/2) * 2 = 20 * 2
This simplifies to:
5n = 40
The multiplication operation serves as a powerful tool for clearing fractions and simplifying equations. By strategically multiplying both sides by the denominator, we eliminate the fraction and transform the equation into a more manageable form.
Unveiling the Variable: The Final Resolution
With the variable term now isolated, the final step involves unveiling the value of 'n'. The variable 'n' is currently multiplied by 5. To isolate 'n', we invoke the inverse operation of multiplication, which is division. Dividing both sides of the equation by 5 reveals the solution:
5n / 5 = 40 / 5
Therefore, the solution to the equation is:
n = 8
This final step underscores the power of inverse operations in isolating variables and solving equations. By strategically applying the inverse operation, we effectively undo the multiplication and reveal the value of the unknown variable.
Verifying the Solution: The Pursuit of Accuracy
The pursuit of accuracy is a cornerstone of mathematical practice. To ensure our solution is correct, we substitute n = 8 back into the original equation, 5n/2 + 15 = 35, and verify if the equation holds true:
(5 * 8) / 2 + 15 = 35
Simplifying the equation:
40 / 2 + 15 = 35
20 + 15 = 35
35 = 35
The equation holds true, validating our solution, n = 8. This verification step is crucial in confirming the accuracy of our solution and ensuring that we have arrived at the correct answer.
Conclusion: A Deeper Understanding of Equation Solving
This in-depth analysis of the equation 5n/2 + 15 = 35 has illuminated the strategic steps involved in solving linear equations. By isolating the variable term, eliminating fractions, and employing inverse operations, we can confidently navigate the world of equations. Remember, consistent practice and a deep understanding of the underlying principles are the keys to unlocking mathematical proficiency. Embrace the challenge, persevere in your efforts, and watch your mathematical skills soar.
Having established a solid foundation in solving linear equations, let's elevate our understanding by exploring advanced techniques and real-world applications. This comprehensive guide will not only reinforce your skills in solving equations like 5n/2 + 15 = 35 but also equip you with the tools to tackle more complex mathematical challenges.
Revisiting the Fundamentals: A Concise Recap
Before we delve into advanced techniques, let's revisit the fundamental principles of solving linear equations. A linear equation, characterized by a variable raised to the power of one, can be expressed in the general form ax + b = c, where 'a', 'b', and 'c' are constants, and 'x' is the variable we aim to solve for. The equation 5n/2 + 15 = 35 serves as an excellent example of this fundamental concept.
The core strategy for solving linear equations revolves around isolating the variable term. This involves a series of strategic maneuvers, including:
- Isolating the variable term: Employing inverse operations to eliminate constant terms and isolate the term containing the variable.
- Eliminating fractions: Multiplying both sides of the equation by the denominator to clear fractions and simplify the equation.
- Solving for the variable: Dividing both sides of the equation by the coefficient of the variable to unveil its value.
- Verifying the solution: Substituting the calculated value back into the original equation to ensure its accuracy.
Advanced Techniques: Unlocking Complex Equations
While the fundamental techniques are effective for solving basic linear equations, more complex scenarios may require advanced strategies. Let's explore some of these techniques:
1. Combining Like Terms
In equations with multiple terms involving the variable or constant terms, combining like terms simplifies the equation. For example, in the equation 2x + 3x + 5 = 15, we can combine 2x and 3x to obtain 5x + 5 = 15.
2. Distributive Property
The distributive property allows us to expand expressions involving parentheses. For example, in the equation 2(x + 3) = 10, we distribute the 2 to both terms inside the parentheses, resulting in 2x + 6 = 10.
3. Solving Equations with Variables on Both Sides
Equations with variables on both sides require a strategic approach to isolate the variable. This involves moving all terms containing the variable to one side of the equation and all constant terms to the other side. For example, in the equation 3x + 5 = x + 9, we can subtract x from both sides and subtract 5 from both sides to obtain 2x = 4.
4. Solving Multi-Step Equations
Many equations require a combination of these techniques to solve. Multi-step equations involve a sequence of operations, such as combining like terms, distributing, and isolating the variable, to arrive at the solution.
Real-World Applications: Connecting Equations to Life
Linear equations are not merely abstract mathematical concepts; they are powerful tools that can be applied to solve a wide range of real-world problems. Let's explore some examples:
1. Calculating Costs and Quantities
Linear equations can be used to calculate the cost of items, the number of items needed, or the total cost of a purchase. For example, if a store sells apples for $2 per pound and you have $10, you can use the equation 2x = 10 to determine that you can buy 5 pounds of apples.
2. Determining Distances and Speeds
Linear equations play a crucial role in calculating distances, speeds, and times. For example, if a car travels at a constant speed of 60 miles per hour, the equation d = 60t can be used to calculate the distance traveled in a given time.
3. Modeling Financial Situations
Linear equations are frequently used to model financial situations, such as calculating interest earned, loan payments, or investment growth. For example, the equation A = P(1 + rt) can be used to calculate the future value of an investment, where A is the future value, P is the principal, r is the interest rate, and t is the time period.
4. Solving Mixture Problems
Mixture problems, involving combining different substances with varying concentrations, can be solved using linear equations. For example, if you need to create a 10% saline solution by mixing a 5% saline solution and a 20% saline solution, you can use linear equations to determine the amount of each solution needed.
Conclusion: Mastering the Art of Equation Solving
This exploration of advanced techniques and real-world applications has broadened our understanding of linear equations. By mastering these concepts, you are equipped to tackle a wider range of mathematical challenges and apply your skills to solve practical problems. Remember, consistent practice, a deep understanding of the underlying principles, and a willingness to explore new techniques are the keys to unlocking mathematical mastery. Embrace the challenge, persevere in your efforts, and watch your mathematical prowess flourish.
This article serves as a comprehensive guide to solving the linear equation 5n/2 + 15 = 35. We will cover everything from the fundamental concepts to advanced techniques, ensuring a thorough understanding of the solution process. Whether you are a student learning algebra or simply looking to refresh your skills, this guide will provide the necessary tools and insights to confidently tackle similar equations.
I. Introduction to Linear Equations
What are Linear Equations?
Linear equations are algebraic equations in which the highest power of the variable is 1. They can be written in the general form ax + b = c, where:
xis the variableais the coefficient of the variablebandcare constants
Our equation, 5n/2 + 15 = 35, fits this form, making it a linear equation.
Why are Linear Equations Important?
Linear equations are fundamental in mathematics and have numerous applications in various fields, including:
- Science
- Engineering
- Economics
- Computer science
Understanding how to solve them is crucial for problem-solving and analytical thinking.
II. Solving the Equation 5n/2 + 15 = 35: A Step-by-Step Approach
Step 1: Isolate the Term with the Variable
To isolate the term with the variable (5n/2), we need to eliminate the constant term (+15) on the left side of the equation. We do this by subtracting 15 from both sides:
5n/2 + 15 - 15 = 35 - 15
This simplifies to:
5n/2 = 20
Step 2: Eliminate the Fraction
To eliminate the fraction, we multiply both sides of the equation by the denominator (2):
(5n/2) * 2 = 20 * 2
This simplifies to:
5n = 40
Step 3: Solve for the Variable
To solve for n, we divide both sides of the equation by the coefficient of n (5):
5n / 5 = 40 / 5
This gives us:
n = 8
Step 4: Verify the Solution
To verify our solution, we substitute n = 8 back into the original equation:
5(8)/2 + 15 = 35
20 + 15 = 35
35 = 35
The equation holds true, so our solution n = 8 is correct.
III. Advanced Techniques for Solving Linear Equations
A. Equations with Variables on Both Sides
When equations have variables on both sides, we need to gather the variable terms on one side and the constants on the other side. For example:
3x + 5 = x + 9
Subtract x from both sides:
2x + 5 = 9
Subtract 5 from both sides:
2x = 4
Divide by 2:
x = 2
B. Equations with Parentheses
When equations have parentheses, we use the distributive property to eliminate them. For example:
2(x + 3) = 10
Distribute the 2:
2x + 6 = 10
Subtract 6 from both sides:
2x = 4
Divide by 2:
x = 2
C. Multi-Step Equations
Many equations require multiple steps to solve. The key is to follow the order of operations (PEMDAS/BODMAS) in reverse to isolate the variable.
IV. Real-World Applications of Linear Equations
Linear equations are used to model many real-world situations. Here are a few examples:
A. Calculating Costs
If a store sells apples for $2 per pound, the equation 2x = C can be used to calculate the cost (C) for x pounds of apples.
B. Determining Distances
If a car travels at a constant speed of 60 miles per hour, the equation d = 60t can be used to calculate the distance (d) traveled in t hours.
C. Solving Mixture Problems
Linear equations can be used to solve mixture problems, such as determining how much of a solution with a certain concentration is needed to create a desired concentration.
V. Conclusion
Solving linear equations is a fundamental skill in mathematics. This guide has provided a comprehensive approach, from basic concepts to advanced techniques and real-world applications. By following the steps outlined and practicing regularly, you can develop confidence and proficiency in solving linear equations.
In this guide, we will thoroughly explore the process of solving the linear equation 5n/2 + 15 = 35. We'll break down each step, provide clear explanations, and offer practice problems to solidify your understanding. Whether you're a student learning algebra or someone seeking to enhance their math skills, this resource is designed to equip you with the knowledge and confidence to tackle linear equations effectively.
Understanding the Basics: What is a Linear Equation?
Before diving into the solution, let's define what a linear equation is. A linear equation is an algebraic equation where the highest power of the variable is one. It can be generally represented in the form ax + b = c, where 'a', 'b', and 'c' are constants, and 'x' is the variable we aim to find. The equation 5n/2 + 15 = 35 fits this definition, making it a linear equation.
Solving 5n/2 + 15 = 35: A Step-by-Step Approach
Now, let's solve the equation 5n/2 + 15 = 35 step-by-step:
Step 1: Isolate the Variable Term
The first step is to isolate the term containing the variable 'n'. In this case, the variable term is 5n/2. To isolate it, we need to eliminate the constant term, which is 15. We can do this by subtracting 15 from both sides of the equation:
5n/2 + 15 - 15 = 35 - 15
Simplifying, we get:
5n/2 = 20
Step 2: Eliminate the Fraction
Next, we need to eliminate the fraction. The variable term is currently divided by 2. To undo this division, we multiply both sides of the equation by 2:
(5n/2) * 2 = 20 * 2
Simplifying, we get:
5n = 40
Step 3: Solve for the Variable
Now, we have 5n = 40. To solve for 'n', we need to isolate it. Since 'n' is being multiplied by 5, we divide both sides of the equation by 5:
5n / 5 = 40 / 5
Simplifying, we get:
n = 8
Step 4: Verify the Solution
It's always a good practice to verify your solution by substituting it back into the original equation. Let's substitute n = 8 into 5n/2 + 15 = 35:
5(8)/2 + 15 = 35
40/2 + 15 = 35
20 + 15 = 35
35 = 35
The equation holds true, so our solution n = 8 is correct.
Advanced Techniques: Solving More Complex Linear Equations
Now that we've mastered the basic steps, let's explore some advanced techniques for solving more complex linear equations:
1. Equations with Variables on Both Sides
Some equations have variables on both sides of the equation. In this case, we need to gather the variable terms on one side and the constant terms on the other side. For example:
3x + 5 = x + 9
Subtract x from both sides:
2x + 5 = 9
Subtract 5 from both sides:
2x = 4
Divide by 2:
x = 2
2. Equations with Parentheses
If an equation has parentheses, we need to use the distributive property to eliminate them. For example:
2(x + 3) = 10
Distribute the 2:
2x + 6 = 10
Subtract 6 from both sides:
2x = 4
Divide by 2:
x = 2
3. Multi-Step Equations
Many equations require a combination of steps to solve. The key is to follow the order of operations (PEMDAS/BODMAS) in reverse to isolate the variable.
Real-World Applications of Linear Equations
Linear equations have numerous applications in real-world scenarios. Here are a few examples:
1. Calculating Costs
If a store sells pens for $1.50 each, the equation 1. 50x = C can be used to calculate the cost (C) for x pens.
2. Determining Distances
If a train travels at a constant speed of 80 miles per hour, the equation d = 80t can be used to calculate the distance (d) traveled in t hours.
3. Solving Mixture Problems
Linear equations can be used to solve mixture problems, such as determining how much of two different solutions needs to be mixed to create a solution with a specific concentration.
Practice Problems
To solidify your understanding, try solving these practice problems:
- 4x - 7 = 9
- 2(y + 5) = 16
- 5z + 3 = 2z + 12
- 3p/4 - 2 = 7
- 6(q - 1) = 3q + 9
Conclusion
Mastering linear equations is a crucial step in developing your algebraic skills. By following the steps outlined in this guide and practicing regularly, you can build confidence and proficiency in solving linear equations. Remember, persistence and practice are key to success in mathematics!
