Solving For Y: A Step-by-Step Guide To Isolate The Variable

by Alex Johnson 60 views

In mathematics, solving for a variable is a fundamental skill. This article will walk you through the process of solving a specific equation for y. We'll break down each step, ensuring you understand the logic and techniques involved. This comprehensive guide will not only help you solve this particular equation but also equip you with the skills to tackle similar problems confidently. Understanding how to isolate a variable is crucial in various mathematical and scientific fields, making this a valuable skill to master.

Understanding the Equation

Before diving into the solution, let's first understand the equation we are dealing with:

โˆ’34=โˆ’14y+73-\frac{3}{4}=-\frac{1}{4} y+\frac{7}{3}

This equation involves fractions, a variable (y), and basic arithmetic operations. To solve for y, our goal is to isolate it on one side of the equation. This means we need to get y by itself, with no other terms or coefficients attached to it. The key to achieving this is to use inverse operations, which essentially undo the operations in the equation. Remember, whatever operation you perform on one side of the equation, you must also perform on the other side to maintain the equality.

The Importance of Isolating Variables

In algebra and beyond, the ability to isolate a variable is paramount. Whether you're dealing with linear equations, quadratic equations, or systems of equations, the core principle remains the same: isolate the variable you're trying to solve for. This skill extends to various applications in physics, engineering, economics, and computer science, where mathematical models often need to be manipulated to find specific values or relationships. By mastering the techniques discussed here, you'll be well-prepared for more advanced mathematical concepts and real-world problem-solving scenarios.

Step 1: Isolate the Term with 'y'

The first step in solving for y is to isolate the term that contains y, which in this case is โˆ’14y-\frac{1}{4}y. To do this, we need to eliminate the 73\frac{7}{3} term on the right side of the equation. We can achieve this by subtracting 73\frac{7}{3} from both sides of the equation. This ensures we maintain the balance of the equation. Remember, whatever operation we perform on one side, we must perform on the other.

โˆ’34โˆ’73=โˆ’14y+73โˆ’73-\frac{3}{4} - \frac{7}{3} = -\frac{1}{4} y + \frac{7}{3} - \frac{7}{3}

Simplifying the right side, 73โˆ’73\frac{7}{3} - \frac{7}{3} cancels out, leaving us with:

โˆ’34โˆ’73=โˆ’14y-\frac{3}{4} - \frac{7}{3} = -\frac{1}{4} y

Now, we need to simplify the left side by subtracting the fractions. To do this, we need a common denominator. The least common multiple of 4 and 3 is 12. So, we'll convert both fractions to have a denominator of 12:

โˆ’34โ‹…33=โˆ’912-\frac{3}{4} \cdot \frac{3}{3} = -\frac{9}{12}

โˆ’73โ‹…44=โˆ’2812-\frac{7}{3} \cdot \frac{4}{4} = -\frac{28}{12}

Now we can substitute these equivalent fractions back into the equation:

โˆ’912โˆ’2812=โˆ’14y-\frac{9}{12} - \frac{28}{12} = -\frac{1}{4} y

Combining the fractions on the left side gives us:

โˆ’3712=โˆ’14y-\frac{37}{12} = -\frac{1}{4} y

Why Isolating Terms is Crucial

The process of isolating terms is a cornerstone of algebraic manipulation. By strategically adding, subtracting, multiplying, or dividing terms on both sides of an equation, we can systematically isolate the variable we're interested in. This step is not only essential for solving equations but also for rearranging formulas in physics, chemistry, and other sciences. Understanding this fundamental principle allows you to solve a wide range of problems efficiently and accurately.

Step 2: Eliminate the Coefficient of 'y'

Now that we have โˆ’3712=โˆ’14y-\frac{37}{12} = -\frac{1}{4} y, our next step is to eliminate the coefficient of y, which is โˆ’14-\frac{1}{4}. To do this, we can multiply both sides of the equation by the reciprocal of โˆ’14-\frac{1}{4}, which is -4. Multiplying by the reciprocal will effectively cancel out the coefficient, leaving y isolated.

โˆ’4โ‹…(โˆ’3712)=โˆ’4โ‹…(โˆ’14y)-4 \cdot \left(-\frac{37}{12}\right) = -4 \cdot \left(-\frac{1}{4} y\right)

On the right side, โˆ’4โ‹…(โˆ’14)-4 \cdot \left(-\frac{1}{4}\right) equals 1, so we have:

โˆ’4โ‹…(โˆ’3712)=y-4 \cdot \left(-\frac{37}{12}\right) = y

Now, let's simplify the left side. We can write -4 as โˆ’41-\frac{4}{1} and multiply the fractions:

โˆ’41โ‹…(โˆ’3712)=y-\frac{4}{1} \cdot \left(-\frac{37}{12}\right) = y

Multiplying the numerators and denominators gives us:

14812=y\frac{148}{12} = y

We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4:

148รท412รท4=y\frac{148 \div 4}{12 \div 4} = y

373=y\frac{37}{3} = y

So, we have found that y equals 373\frac{37}{3}.

The Power of Inverse Operations

This step highlights the power of inverse operations in solving equations. By multiplying by the reciprocal, we effectively "undid" the multiplication by โˆ’14-\frac{1}{4}. This principle of using inverse operations is fundamental to solving all types of equations, from simple linear equations to more complex polynomial and trigonometric equations. Understanding how to apply inverse operations is a crucial skill for any student of mathematics.

Step 3: Verify the Solution

To ensure our solution is correct, it's always a good practice to verify it by substituting the value of y back into the original equation. This step helps catch any potential errors made during the solving process. Our solution is y=373y = \frac{37}{3}, so we'll substitute this value into the original equation:

โˆ’34=โˆ’14y+73-\frac{3}{4}=-\frac{1}{4} y+\frac{7}{3}

โˆ’34=โˆ’14(373)+73-\frac{3}{4}=-\frac{1}{4} \left(\frac{37}{3}\right)+\frac{7}{3}

Now, let's simplify the right side:

โˆ’34=โˆ’3712+73-\frac{3}{4}=-\frac{37}{12}+\frac{7}{3}

To add the fractions on the right side, we need a common denominator, which is 12. So, we'll convert 73\frac{7}{3} to have a denominator of 12:

73โ‹…44=2812\frac{7}{3} \cdot \frac{4}{4} = \frac{28}{12}

Now we can substitute this equivalent fraction back into the equation:

โˆ’34=โˆ’3712+2812-\frac{3}{4}=-\frac{37}{12}+\frac{28}{12}

Combining the fractions on the right side gives us:

โˆ’34=โˆ’912-\frac{3}{4}=-\frac{9}{12}

Now, let's simplify โˆ’912-\frac{9}{12} by dividing both the numerator and denominator by their greatest common divisor, which is 3:

โˆ’9รท312รท3=โˆ’34-\frac{9 \div 3}{12 \div 3} = -\frac{3}{4}

So, we have:

โˆ’34=โˆ’34-\frac{3}{4}=-\frac{3}{4}

Since the left side equals the right side, our solution y=373y = \frac{37}{3} is correct.

The Importance of Verification

Verifying your solution is a crucial step in the problem-solving process. It's a way to double-check your work and ensure that you haven't made any errors along the way. By substituting your solution back into the original equation, you can confirm that it satisfies the equation's conditions. This practice not only enhances your confidence in your answer but also helps you develop a deeper understanding of the problem and the solution process.

Conclusion

In this article, we've walked through the step-by-step process of solving the equation โˆ’34=โˆ’14y+73-\frac{3}{4}=-\frac{1}{4} y+\frac{7}{3} for y. We've covered how to isolate the term with y, eliminate its coefficient, and verify the solution. By understanding these techniques, you can confidently solve similar equations and tackle more complex mathematical problems. Remember, the key is to break down the problem into smaller, manageable steps and apply the appropriate inverse operations. Mastering these fundamental skills will undoubtedly benefit you in your mathematical journey.

For further learning and practice on solving equations, you can explore resources like Khan Academy's Algebra section.