Mastering Linear Equations: Solve (1/7)x + (1/8)y = 1

by Admin 54 views
Mastering Linear Equations: Solve (1/7)x + (1/8)y = 1Looking to **master linear equations**? Well, you've landed in the right spot, guys! Today, we're diving deep into *solving a system of linear equations* that might look a little intimidating at first glance because, let's be honest, those fractions can sometimes throw us for a loop. But don't you worry, by the time we're done, you'll feel like a pro at tackling these kinds of challenges. We're going to break down the problem: given the equations $\frac{1}{7}x + \frac{1}{8}y = 1$ and $\frac{1}{4}x - \frac{1}{6}y = 1$, our ultimate goal is to find the exact values for *x* and *y* that make both statements true simultaneously. Think of it like a treasure hunt where *x* and *y* are our hidden gems! Understanding how to **solve systems of linear equations** is a fundamental skill in mathematics, crucial for everything from basic algebra to advanced calculus, and even practical applications in engineering, economics, and science. It's not just about getting the right answer; it's about understanding the underlying principles and developing a systematic approach. Many students find fractions daunting, leading them to avoid problems like this, but we're here to show you that with the right strategy, they're nothing to fear. We'll walk through the *elimination method*, which is incredibly powerful, especially when you have coefficients that aren't nice, round integers. Our friendly, step-by-step guide will cover everything from clearing those pesky fractions using the **Least Common Denominator (LCD)** to performing the final checks on our solution. So, grab a coffee, get comfortable, and let's conquer this system together! You'll gain valuable insights and boost your confidence in *algebraic problem-solving* along the way. Get ready to transform those fractional equations into something much more manageable and ultimately find our elusive *x* and *y* values. The journey to **mastering linear equations** starts right here, right now, with this specific problem as our exciting playground.## Tackling the Fractional Challenge: Simplifying Our EquationsFirst up, we need to address the elephant in the room: *fractions*. Seriously, who likes dealing with fractions more than they have to? Not many of us, I'd bet! Our first equation is $\frac{1}{7}x + \frac{1}{8}y = 1$. The best way to make this equation friendlier is to *eliminate the fractions* entirely. To do this, we need to find the **Least Common Denominator (LCD)** of all the denominators present in this equation. In this case, our denominators are 7 and 8. What's the smallest number that both 7 and 8 divide into evenly? Since 7 is a prime number and 8 is $2^3$, they don't share any common factors other than 1. So, their LCD is simply their product: $7 \times 8 = 56$. This **LCD** is our magic multiplier! We're going to multiply *every single term* in the entire equation by 56. Let's see how that works: $56 \times (\frac{1}{7}x) + 56 \times (\frac{1}{8}y) = 56 \times 1$. When we multiply $56 \times \frac{1}{7}x$, the 56 and 7 simplify, leaving us with $8x$. Similarly, $56 \times \frac{1}{8}y$ simplifies to $7y$. And of course, $56 \times 1$ is just 56. So, our new, much cleaner equation, let's call it Equation 1', becomes $8x + 7y = 56$. See? Not so bad, right? This strategy of using the **Least Common Denominator** is a crucial technique for *solving linear equations with fractions*, turning a potentially messy problem into a straightforward one. It's all about making the algebra simpler for ourselves and reducing the chances of computational errors. Always remember to multiply *every term* on both sides of the equation by the LCD; it's a common mistake to forget one term, which can throw off your entire solution. This simple step is fundamental to **solving systems of linear equations** efficiently, particularly when fractions are involved, as it sets the stage for easier subsequent steps, like using the elimination or substitution method.Now, let's apply the same fantastic strategy to our second equation: $\frac{1}{4}x - \frac{1}{6}y = 1$. Just like before, we want to *get rid of those fractions*. Look at the denominators: 4 and 6. What's the **Least Common Denominator (LCD)** for 4 and 6? If you list out multiples, you'll find that 12 is the smallest number that both 4 and 6 divide into evenly ($4 \times 3 = 12$ and $6 \times 2 = 12$). So, 12 is our multiplier for this equation! We're going to multiply *every term* in this equation by 12. Let's break it down: $12 \times (\frac{1}{4}x) - 12 \times (\frac{1}{6}y) = 12 \times 1$. For the first term, $12 \times \frac{1}{4}x$, the 12 and 4 simplify, giving us $3x$. For the second term, $-12 \times \frac{1}{6}y$, the 12 and 6 simplify, resulting in $-2y$. And naturally, $12 \times 1$ is just 12. So, our second new, fraction-free equation, Equation 2', is $3x - 2y = 12$. Voila! No more fractions to worry about. This transformation is incredibly powerful for *solving systems of linear equations* because it simplifies the subsequent algebra tremendously. We've gone from a system that looked a bit scary with all its fractions to a new system that is perfectly clean and ready to be solved using standard methods. Our simplified system now looks like this:Equation 1': $8x + 7y = 56$Equation 2': $3x - 2y = 12$This process of clearing fractions is not just a neat trick; it's a **fundamental step** in making complex *algebraic equations* manageable. By carefully finding the **LCD** for each equation and multiplying correctly, we've set ourselves up for success. This preparation ensures that the core task of **solving for x and y** becomes a matter of straightforward arithmetic and basic algebra, rather than wrestling with fractional arithmetic. It highlights the importance of strategic simplification in *mathematical problem-solving* and is a technique you'll use repeatedly.## The Elimination Method: Strategizing Our AttackAlright, guys, now that we've got our super-clean, fraction-free system: $8x + 7y = 56$ (Equation 1') and $3x - 2y = 12$ (Equation 2'), it's time to unleash the power of the *elimination method*. This method is brilliant because it allows us to 'eliminate' one of the variables by adding or subtracting the two equations, leaving us with a single equation that has only one variable. To do this, we need the coefficients of either *x* or *y* to be opposites. Looking at our system, it seems a bit easier to target *y* for elimination because we have a +7y in Equation 1' and a -2y in Equation 2'. The opposite signs are already a good start! To make their coefficients opposites, we need to find the **Least Common Multiple (LCM)** of 7 and 2. The smallest number that both 7 and 2 divide into is 14. So, our goal is to turn one `y` term into +14y and the other into -14y.To achieve this, we'll multiply Equation 1' by 2 (to get +14y) and Equation 2' by 7 (to get -14y). Remember, when you multiply an equation, you have to multiply *every single term* on both sides to keep the equation balanced. Let's do it carefully:Multiply Equation 1' by 2: $2 \times (8x + 7y = 56)$ gives us $16x + 14y = 112$. Let's call this Equation 3.Multiply Equation 2' by 7: $7 \times (3x - 2y = 12)$ gives us $21x - 14y = 84$. Let's call this Equation 4.Now, check out our new system:Equation 3: $16x + 14y = 112$Equation 4: $21x - 14y = 84$Notice how the `y` terms, +14y and -14y, are now perfect opposites? This is exactly what we wanted! This strategic step is the heart of the *elimination method* and is incredibly effective for **solving systems of linear equations**. By carefully choosing our multipliers based on the **LCM**, we ensure that one variable will vanish when we combine the equations. This preparation is paramount for ensuring a smooth path to finding the individual values of *x* and *y*, showcasing the elegance and efficiency of algebraic manipulation. It highlights the importance of looking ahead in your problem-solving process to simplify the next steps, a key aspect of becoming proficient in *mathematics*.## Solving for Our Variables: The Final PushWith our equations expertly prepped – Equation 3: $16x + 14y = 112$ and Equation 4: $21x - 14y = 84$ – we are now at the most exciting part of the *elimination method*: combining them! Since our `y` terms have opposite signs and identical coefficients (+14y and -14y), we can simply *add* Equation 3 and Equation 4 together. This addition will cause the `y` terms to cancel each other out, leaving us with an equation containing only *x*.Let's add them term by term:$(16x + 21x) + (14y - 14y) = (112 + 84)$37x + 0y = 196$37x = 196$Boom! Just like that, the `y` disappeared, and we're left with a simple linear equation in *x*. Now, to find *x*, all we need to do is isolate it by dividing both sides of the equation by 37:$\frac{37x}{37} = \frac{196}{37}$So, $x = \frac{196}{37}$.This value might look a little 'unclean' because it's a fraction, but hey, not all solutions are neat integers, and that's perfectly okay in *mathematics*! The key is that we've found an exact value for *x* that satisfies our conditions. This step clearly demonstrates the power of the *elimination method* for **solving systems of linear equations**. By carefully manipulating the original equations to create opposite coefficients, we were able to reduce a two-variable problem to a much simpler one-variable problem. This systematic approach not only provides the correct answer but also builds a strong foundation for tackling more complex algebraic challenges, reinforcing the principles of *equation solving*. The careful execution of this step is what drives us closer to our complete solution for *x* and *y*, showcasing precision in algebraic manipulation.Now that we've successfully found our value for *x*, which is $x = \frac{196}{37}$, our next mission is to find *y*. The easiest way to do this is by taking our *x* value and *substituting* it back into one of our *simplified, fraction-free equations*. Why simplified ones? Because they're just cleaner and involve less arithmetic, reducing the chances of errors. Let's pick Equation 2', which was $3x - 2y = 12$, because it involves smaller numbers than Equation 1' (8x + 7y = 56) and might make the calculations slightly more manageable.Substitute $x = \frac{196}{37}$ into Equation 2':$3\left(\frac{196}{37}\right) - 2y = 12$First, let's multiply 3 by $\frac{196}{37}$:$\frac{3 \times 196}{37} - 2y = 12$\frac{588}{37} - 2y = 12$Now, we need to isolate the *y* term. Let's move the fractional term to the right side of the equation by subtracting $\frac{588}{37}$ from both sides:$-2y = 12 - \frac{588}{37}$To combine the terms on the right side, we need a common denominator, which is 37. Convert 12 to a fraction with a denominator of 37: $12 = \frac{12 \times 37}{37} = \frac{444}{37}$.$-2y = \frac{444}{37} - \frac{588}{37}$-2y = \frac{444 - 588}{37}$-2y = \frac{-144}{37}$Finally, to solve for *y*, we need to divide both sides by -2 (or multiply by $-\frac{1}{2}$):$y = \frac{-144}{37} \div (-2)$y = \frac{-144}{37} \times \left(-\frac{1}{2}\right)$The negative signs cancel out, and 144 divided by 2 is 72:$y = \frac{72}{37}$And there you have it, guys! We've found both our variables: $x = \frac{196}{37}$ and $y = \frac{72}{37}$. This process of *substitution* is critical in **solving systems of linear equations** once one variable is known. It's a precise application of algebraic rules to ensure we arrive at the correct value for the second variable. This entire sequence, from clearing fractions to elimination and finally substitution, truly showcases the systematic nature of *mathematical problem-solving* and prepares you for more advanced *algebraic manipulations*.## The Confidence Check: Verifying Our SolutionAlright, my friends, we've done all the heavy lifting! We've found $x = \frac{196}{37}$ and $y = \frac{72}{37}$. But in *mathematics*, especially when dealing with complex problems like **solving systems of linear equations** with fractions, the real mark of a pro is *verification*. It's not enough to just get an answer; we need to be absolutely sure that our solution works for *both* of the *original equations*. This step is crucial for building confidence and catching any potential arithmetic errors we might have made along the way. Let's plug our values back into the *original* equations with fractions:Equation 1: $\frac{1}{7}x + \frac{1}{8}y = 1$Substitute *x* and *y*:$\frac{1}{7}\left(\frac{196}{37}\right) + \frac{1}{8}\left(\frac{72}{37}\right) = 1$Let's simplify each term:$\frac{196}{7 \times 37} + \frac{72}{8 \times 37} = 1$Recognize that $196 = 7 \times 28$ and $72 = 8 \times 9$:$\frac{28}{37} + \frac{9}{37} = 1$Add the fractions:$\frac{28 + 9}{37} = 1$\frac{37}{37} = 1$1 = 1$Fantastic! Equation 1 holds true with our values. That's a huge sigh of relief, but we're not done yet! We need to check the second original equation as well to fully **verify our solution**.Equation 2: $\frac{1}{4}x - \frac{1}{6}y = 1$Substitute *x* and *y*:$\frac{1}{4}\left(\frac{196}{37}\right) - \frac{1}{6}\left(\frac{72}{37}\right) = 1$Simplify each term:$\frac{196}{4 \times 37} - \frac{72}{6 \times 37} = 1$Recognize that $196 = 4 \times 49$ and $72 = 6 \times 12$:$\frac{49}{37} - \frac{12}{37} = 1$Subtract the fractions:$\frac{49 - 12}{37} = 1$\frac{37}{37} = 1$1 = 1$Absolutely perfect! Both original equations are satisfied by our calculated values for *x* and *y*. This thorough *verification process* confirms that our solution $x = \frac{196}{37}$ and $y = \frac{72}{37}$ is correct. It's the final, crucial step in **solving systems of linear equations** and demonstrates a complete understanding of the problem. This attention to detail is what makes your mathematical work truly stand out and solidifies your mastery of *algebraic problem-solving*. Always remember to take this extra step – it's a small investment of time that pays off big in confidence and accuracy! So, the final answer is $x = \frac{196}{37}$ and $y = \frac{72}{37}$. Congratulations on mastering this complex system! This process provides immense value by ensuring the correctness of your results and reinforcing the logical flow of *mathematical deduction*. This entire journey, from simplifying fractions to using the *elimination method* and finally verifying, showcases a robust approach to *algebra*. It's a critical skill for any student aiming to excel in *mathematics*.