Solving For X When 5x + 7 Is One-Third Of 61 - 10x

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In the realm of mathematics, solving equations is a fundamental skill. This article delves into a specific algebraic problem where we aim to find the value of the variable x that satisfies a given condition. The problem states: for what value of the variable is the value of the expression $5x + 7$ one-third of the value of the expression $61 - 10x$? This requires us to translate the word problem into a mathematical equation and then solve for x. Let's embark on this mathematical journey step by step.

Translating the Word Problem into an Equation

To begin, we need to express the given condition as a mathematical equation. The phrase "one-third of" implies division by 3. Therefore, we can write the equation as follows:

5x+7=13(61−10x)5x + 7 = \frac{1}{3}(61 - 10x)

This equation represents the core of our problem. It states that the value of the expression 5x + 7 is equal to one-third of the value of the expression 61 - 10x. Now, our task is to solve this equation for x.

Solving the Equation for x

To solve the equation, we need to isolate x on one side. The first step is to eliminate the fraction by multiplying both sides of the equation by 3:

3(5x+7)=61−10x3(5x + 7) = 61 - 10x

Next, we distribute the 3 on the left side:

15x+21=61−10x15x + 21 = 61 - 10x

Now, we want to gather the terms containing x on one side and the constant terms on the other side. To do this, we add 10x to both sides:

15x+10x+21=6115x + 10x + 21 = 61

25x+21=6125x + 21 = 61

Next, we subtract 21 from both sides:

25x=61−2125x = 61 - 21

25x=4025x = 40

Finally, we divide both sides by 25 to solve for x:

x=4025x = \frac{40}{25}

Simplifying the fraction, we get:

x=85x = \frac{8}{5}

Therefore, the value of x that satisfies the given condition is 8/5.

Verifying the Solution

To ensure our solution is correct, we can substitute x = 8/5 back into the original equation and verify that both sides are equal. Let's do that:

Left side:

5x+7=5(85)+7=8+7=155x + 7 = 5(\frac{8}{5}) + 7 = 8 + 7 = 15

Right side:

13(61−10x)=13(61−10(85))=13(61−16)=13(45)=15\frac{1}{3}(61 - 10x) = \frac{1}{3}(61 - 10(\frac{8}{5})) = \frac{1}{3}(61 - 16) = \frac{1}{3}(45) = 15

Since both sides are equal to 15, our solution x = 8/5 is correct.

Alternative Approaches to Solving the Equation

While we have solved the equation using a standard algebraic approach, there might be alternative methods to arrive at the same solution. For instance, one could choose to distribute the 1/3 on the right side of the equation first. This would lead to:

5x+7=613−103x5x + 7 = \frac{61}{3} - \frac{10}{3}x

Then, we could add $\frac{10}{3}x$ to both sides:

5x+103x+7=6135x + \frac{10}{3}x + 7 = \frac{61}{3}

Combining the x terms:

253x+7=613\frac{25}{3}x + 7 = \frac{61}{3}

Subtracting 7 from both sides (note that 7 is equivalent to $\frac{21}{3}$):

253x=613−213\frac{25}{3}x = \frac{61}{3} - \frac{21}{3}

253x=403\frac{25}{3}x = \frac{40}{3}

Finally, multiplying both sides by $\frac{3}{25}$:

x=403â‹…325=4025=85x = \frac{40}{3} \cdot \frac{3}{25} = \frac{40}{25} = \frac{8}{5}

This alternative approach confirms our previous solution, showcasing the flexibility in solving algebraic equations.

Importance of Verification in Mathematical Problem Solving

The verification step in solving mathematical problems is crucial. It serves as a safeguard against potential errors made during the solution process. By substituting the obtained value back into the original equation, we can confirm whether the solution satisfies the given condition. This practice not only enhances our confidence in the answer but also deepens our understanding of the problem itself. Furthermore, if the verification fails, it indicates that there might be an error in our calculations, prompting us to revisit the steps and identify the mistake.

Real-World Applications of Algebraic Equations

Algebraic equations are not just abstract mathematical concepts; they have numerous applications in real-world scenarios. From calculating financial investments to determining the trajectory of a projectile, algebraic equations provide a powerful tool for modeling and solving problems in various fields. In physics, they are used to describe motion, forces, and energy. In engineering, they are essential for designing structures, circuits, and systems. In economics, they help analyze market trends and predict economic behavior. The ability to formulate and solve algebraic equations is therefore a valuable skill in many professions and aspects of life.

Conclusion: Mastering Algebraic Problem Solving

In conclusion, we have successfully solved the problem of finding the value of x for which the expression 5x + 7 is one-third of the expression 61 - 10x. The solution, x = 8/5, was obtained by translating the word problem into a mathematical equation, solving the equation using algebraic techniques, and verifying the solution. We also explored an alternative approach to solving the equation, highlighting the versatility of algebraic methods. The importance of verification in ensuring the correctness of the solution was emphasized, and the real-world applications of algebraic equations were discussed. Mastering algebraic problem solving is a fundamental skill in mathematics and has far-reaching implications in various fields. Through practice and a solid understanding of algebraic principles, one can confidently tackle a wide range of mathematical challenges. Understanding these concepts is crucial for success in mathematics and related fields. This exercise underscores the power and elegance of algebra in solving practical problems.

In this detailed exploration, we aim to dissect and solve the mathematical problem presented: determining the value of the variable x for which the expression 5x + 7 equals one-third of the expression 61 - 10x. This problem, rooted in algebraic principles, exemplifies a common type of equation-solving challenge encountered in mathematics. To effectively address this, we will systematically break down the problem, translate it into a mathematical equation, and employ algebraic techniques to isolate and solve for x. Furthermore, we will delve into the significance of verifying our solution, ensuring its accuracy and validity. Lastly, we will touch upon the broader implications and real-world applications of such algebraic problem-solving skills.

The Initial Setup: Deconstructing the Problem Statement

The problem statement, "For what value of the variable is the value of the expression 5x + 7 one-third of the value of the expression 61 - 10x?" can initially seem complex. However, the key lies in carefully deconstructing the sentence into its mathematical components. The phrase "the value of the expression 5x + 7" directly refers to the algebraic expression itself. The term "one-third of" implies a fractional relationship, specifically dividing by 3. The expression "the value of the expression 61 - 10x" presents the second algebraic component of the problem. The central question asks for the x value that makes the first expression equal to one-third of the second. This conceptual understanding forms the foundation for translating the problem into a solvable equation.

From Words to Equations: Translating the Problem Mathematically

The transition from a word problem to a mathematical equation is a critical step in problem-solving. In this instance, the verbal statement can be precisely mirrored in algebraic notation. The phrase "5x + 7 is one-third of 61 - 10x" translates directly to the equation:

5x+7=13(61−10x)5x + 7 = \frac{1}{3}(61 - 10x)

This equation encapsulates the core of the problem. It asserts that the expression on the left side (5x + 7) is equivalent to one-third of the expression on the right side (61 - 10x). The equation becomes the focal point for our algebraic manipulation, with the goal of isolating x and determining its value.

The Solution Process: Step-by-Step Algebraic Manipulation

Solving the equation requires a methodical approach, employing algebraic principles to simplify and isolate x. The initial hurdle is dealing with the fraction. To eliminate it, we multiply both sides of the equation by 3:

3(5x+7)=3⋅13(61−10x)3(5x + 7) = 3 \cdot \frac{1}{3}(61 - 10x)

This simplifies to:

15x+21=61−10x15x + 21 = 61 - 10x

Next, we aim to group like terms. To do this, we add 10x to both sides:

15x+10x+21=61−10x+10x15x + 10x + 21 = 61 - 10x + 10x

25x+21=6125x + 21 = 61

Now, we isolate the term with x by subtracting 21 from both sides:

25x+21−21=61−2125x + 21 - 21 = 61 - 21

25x=4025x = 40

Finally, to solve for x, we divide both sides by 25:

25x25=4025\frac{25x}{25} = \frac{40}{25}

x=4025x = \frac{40}{25}

Simplifying the fraction, we find:

x=85x = \frac{8}{5}

Thus, the value of x that satisfies the original condition is 8/5. This result represents the solution to our algebraic problem.

The Importance of Verification: Ensuring Solution Accuracy

The process of verifying a solution is paramount in mathematical problem-solving. It acts as a safeguard against potential errors made during the algebraic manipulation. To verify our solution, x = 8/5, we substitute this value back into the original equation:

5(85)+7=13(61−10(85))5(\frac{8}{5}) + 7 = \frac{1}{3}(61 - 10(\frac{8}{5}))

Evaluating the left side:

5(85)+7=8+7=155(\frac{8}{5}) + 7 = 8 + 7 = 15

Evaluating the right side:

13(61−10(85))=13(61−16)=13(45)=15\frac{1}{3}(61 - 10(\frac{8}{5})) = \frac{1}{3}(61 - 16) = \frac{1}{3}(45) = 15

Since both sides of the equation equal 15, our solution x = 8/5 is verified as correct. This verification step reinforces confidence in our answer and confirms the accuracy of our algebraic process.

Alternative Solution Paths: Flexibility in Algebra

While the previously outlined method provides a clear path to the solution, algebraic problems often offer alternative routes. For instance, instead of initially multiplying both sides by 3, we could distribute the 1/3 on the right side:

5x+7=13(61)−13(10x)5x + 7 = \frac{1}{3}(61) - \frac{1}{3}(10x)

5x+7=613−10x35x + 7 = \frac{61}{3} - \frac{10x}{3}

Then, we could add $\frac{10x}{3}$ to both sides:

5x+10x3+7=6135x + \frac{10x}{3} + 7 = \frac{61}{3}

Combining the x terms:

15x3+10x3+7=613\frac{15x}{3} + \frac{10x}{3} + 7 = \frac{61}{3}

25x3+7=613\frac{25x}{3} + 7 = \frac{61}{3}

Subtracting 7 (or $\frac{21}{3}$) from both sides:

25x3=613−213\frac{25x}{3} = \frac{61}{3} - \frac{21}{3}

25x3=403\frac{25x}{3} = \frac{40}{3}

Finally, multiplying both sides by $\frac{3}{25}$:

x=403â‹…325x = \frac{40}{3} \cdot \frac{3}{25}

x=4025x = \frac{40}{25}

x=85x = \frac{8}{5}

This alternative approach arrives at the same solution, x = 8/5, demonstrating the inherent flexibility within algebraic problem-solving. It underscores that multiple valid methods can lead to the correct answer, fostering a deeper understanding of algebraic principles.

Real-World Relevance: Applications Beyond the Classroom

Algebraic equations are not merely abstract exercises; they are fundamental tools with widespread real-world applications. From finance and engineering to physics and computer science, algebraic principles underpin countless practical scenarios. For example, in finance, equations are used to calculate interest rates, loan payments, and investment returns. In engineering, they are essential for designing structures, analyzing circuits, and optimizing systems. In physics, they describe motion, forces, and energy. In computer science, they are used in algorithm design and data analysis. The ability to formulate and solve algebraic equations is thus a valuable skill across diverse fields, enabling individuals to model, analyze, and solve real-world problems effectively. Mastering these skills opens doors to a wide range of career opportunities and enhances problem-solving capabilities in everyday life. A strong foundation in algebra is therefore crucial for success in numerous academic and professional pursuits.

Synthesizing Knowledge: The Power of Algebraic Thinking

In conclusion, the problem of finding x such that 5x + 7 is one-third of 61 - 10x serves as a microcosm of algebraic problem-solving. By carefully translating the problem statement into an equation, employing step-by-step algebraic manipulation, verifying the solution, and exploring alternative approaches, we have demonstrated the power and versatility of algebraic thinking. Furthermore, by recognizing the real-world relevance of algebraic equations, we underscore the importance of mastering these skills for academic, professional, and everyday success. Algebraic thinking is not just about manipulating symbols; it's about developing a logical and systematic approach to problem-solving, a skill that is invaluable in a wide range of contexts. The ability to think algebraically empowers individuals to tackle complex challenges with confidence and precision. Developing these skills is an investment in one's future, opening doors to new opportunities and enhancing problem-solving capabilities in all aspects of life.

In the realm of algebra, solving linear equations is a fundamental skill. This article focuses on a specific problem that requires us to find the value of the variable x in a linear equation derived from a word problem. The problem states: Determine the value of x for which the expression 5x + 7 is equal to one-third of the expression 61 - 10x. This involves translating the given statement into a mathematical equation and then employing algebraic techniques to isolate x and find its value. Let's delve into the step-by-step solution process.

Transforming Words into a Mathematical Equation

The initial step in solving this problem is to convert the verbal statement into a precise mathematical equation. The phrase "one-third of" signifies division by 3. Therefore, the statement "5x + 7 is equal to one-third of 61 - 10x" can be expressed as the equation:

5x+7=13(61−10x)5x + 7 = \frac{1}{3}(61 - 10x)

This equation forms the basis for our subsequent algebraic manipulations. It establishes a relationship between the two expressions, 5x + 7 and 61 - 10x, stipulating that the former is one-third of the latter. Solving this equation will reveal the value of x that satisfies this condition.

Step-by-Step Solution: Isolating x

To solve the equation, we need to isolate x on one side. The first step is to eliminate the fraction by multiplying both sides of the equation by 3:

3(5x+7)=3⋅13(61−10x)3(5x + 7) = 3 \cdot \frac{1}{3}(61 - 10x)

This simplifies to:

15x+21=61−10x15x + 21 = 61 - 10x

Next, we group the terms containing x on one side and the constant terms on the other. To do this, we add 10x to both sides:

15x+10x+21=61−10x+10x15x + 10x + 21 = 61 - 10x + 10x

25x+21=6125x + 21 = 61

Now, we subtract 21 from both sides to isolate the term with x:

25x+21−21=61−2125x + 21 - 21 = 61 - 21

25x=4025x = 40

Finally, we divide both sides by 25 to solve for x:

25x25=4025\frac{25x}{25} = \frac{40}{25}

x=4025x = \frac{40}{25}

Simplifying the fraction, we obtain:

x=85x = \frac{8}{5}

Therefore, the solution to the equation is x = 8/5. This is the value of x that makes the expression 5x + 7 equal to one-third of the expression 61 - 10x.

Validation of the Solution: Verification Process

To ensure the accuracy of our solution, it is crucial to verify it by substituting x = 8/5 back into the original equation:

5(85)+7=13(61−10(85))5(\frac{8}{5}) + 7 = \frac{1}{3}(61 - 10(\frac{8}{5}))

Evaluating the left side:

5(85)+7=8+7=155(\frac{8}{5}) + 7 = 8 + 7 = 15

Evaluating the right side:

13(61−10(85))=13(61−16)=13(45)=15\frac{1}{3}(61 - 10(\frac{8}{5})) = \frac{1}{3}(61 - 16) = \frac{1}{3}(45) = 15

Since both sides of the equation equal 15, our solution x = 8/5 is confirmed to be correct. This verification step is essential in problem-solving, as it helps identify any potential errors made during the algebraic manipulation process.

An Alternative Approach: Varying the Solution Path

While we have solved the equation using a standard method, alternative approaches can also be employed. For instance, instead of multiplying both sides by 3 initially, we could distribute the 1/3 on the right side of the equation:

5x+7=613−10x35x + 7 = \frac{61}{3} - \frac{10x}{3}

Then, we could add $\frac{10x}{3}$ to both sides:

5x+10x3+7=6135x + \frac{10x}{3} + 7 = \frac{61}{3}

Combining the x terms:

15x3+10x3+7=613\frac{15x}{3} + \frac{10x}{3} + 7 = \frac{61}{3}

25x3+7=613\frac{25x}{3} + 7 = \frac{61}{3}

Subtracting 7 (which is equivalent to $\frac{21}{3}$) from both sides:

25x3=613−213\frac{25x}{3} = \frac{61}{3} - \frac{21}{3}

25x3=403\frac{25x}{3} = \frac{40}{3}

Finally, multiplying both sides by $\frac{3}{25}$:

x=403â‹…325x = \frac{40}{3} \cdot \frac{3}{25}

x=4025x = \frac{40}{25}

x=85x = \frac{8}{5}

This alternative approach yields the same solution, x = 8/5, demonstrating that multiple valid methods can be used to solve algebraic equations. This highlights the flexibility and adaptability inherent in algebraic problem-solving.

The Wide Applicability of Linear Equations

Linear equations are not confined to the realm of textbooks; they are ubiquitous in real-world applications. They are used to model a wide variety of phenomena, from simple relationships to complex systems. In physics, they can describe the motion of objects under constant acceleration. In economics, they can model supply and demand curves. In computer science, they are used in linear programming and optimization problems. The ability to solve linear equations is therefore a valuable skill in numerous fields, enabling individuals to analyze and solve practical problems effectively. Mastering these techniques is essential for success in many academic and professional endeavors. Understanding linear equations is a cornerstone of mathematical literacy and a key to unlocking a deeper understanding of the world around us.

Concluding Thoughts: Mastering the Art of Equation Solving

In summary, we have successfully determined the value of x for which the expression 5x + 7 is one-third of the expression 61 - 10x. The solution, x = 8/5, was obtained by translating the word problem into a mathematical equation, employing algebraic techniques to isolate x, verifying the solution, and exploring an alternative approach. This exercise underscores the importance of a systematic approach to problem-solving and the versatility of algebraic methods. The art of equation solving lies not only in finding the correct answer but also in understanding the underlying principles and developing the ability to apply them in different contexts. Mastering these skills empowers individuals to tackle a wide range of mathematical challenges with confidence and precision. Continuous practice and a deep understanding of algebraic concepts are key to achieving proficiency in equation solving and unlocking its many applications.