Solving Rational Equations: A Step-by-Step Guide

When you encounter a math problem like solving the equation $\frac{8}{5 y}+\frac{8}{15}=\frac{7}{30 y}$, it might seem a bit daunting at first glance, especially with those variables in the denominators. But don't worry! These are called rational equations, and with a systematic approach, you can tackle them effectively. The key here is to eliminate the denominators, making the equation much simpler to solve. We'll walk through this step-by-step, ensuring you understand each part of the process. The first crucial step in solving any rational equation is to identify the least common denominator (LCD) of all the fractions involved. This is the smallest expression that all the denominators can divide into evenly. In our equation, $\frac{8}{5 y}+\frac{8}{15}=\frac{7}{30 y}$, our denominators are $5y$, $15$, and $30y$. To find the LCD, let's break down each denominator into its prime factors. For $5y$, the factors are $5$ and $y$. For $15$, the factors are $3$ and $5$. For $30y$, the factors are $2$, $3$, $5$, and $y$. Now, to find the LCD, we take the highest power of each prime factor that appears in any of the denominators. We have the prime factors $2$, $3$, $5$, and $y$. The highest power of $2$ is $2^1$, the highest power of $3$ is $3^1$, the highest power of $5$ is $5^1$, and the highest power of $y$ is $y^1$. Therefore, the LCD is $2 \times 3 \times 5 \times y = 30y$. This LCD, $30y$, will be our secret weapon to simplify the equation. Once we have the LCD, the next logical step is to multiply every term in the equation by this LCD. This action will effectively clear out all the denominators, transforming the rational equation into a simpler polynomial equation that we already know how to solve. So, let's multiply each term of $\frac{8}{5 y}+\frac{8}{15}=\frac{7}{30 y}$ by $30y$: $(30y) \times \frac{8}{5 y} + (30y) \times \frac{8}{15} = (30y) \times \frac{7}{30 y}$. Now, let's simplify each term. For the first term, $(30y) \times \frac{8}{5 y}$, the $y$ in the numerator and denominator cancel out, and $30$ divided by $5$ is $6$. So, this term becomes $6 \times 8 = 48$. For the second term, $(30y) \times \frac{8}{15}$, the $30$ divided by $15$ is $2$. So, this term becomes $2y \times 8 = 16y$. For the third term, $(30y) \times \frac{7}{30 y}$, both the $30y$ in the numerator and denominator cancel out, leaving us with just $7$. So, the equation transforms from $\frac{8}{5 y}+\frac{8}{15}=\frac{7}{30 y}$ to $48 + 16y = 7$. See? Much simpler! This is now a linear equation, which is straightforward to solve. The goal is to isolate the variable $y$. First, we want to get the term with $y$ by itself on one side of the equation. To do this, we subtract $48$ from both sides: $16y = 7 - 48$. This gives us $16y = -41$. Finally, to find the value of $y$, we divide both sides by the coefficient of $y$, which is $16$: $y = \frac{-41}{16}$. So, the solution to our original rational equation is $y = -\frac{41}{16}$. However, there's a critical final step when solving rational equations that many people forget: checking for extraneous solutions. Extraneous solutions are values of the variable that make one or more of the original denominators equal to zero. If we substitute an extraneous solution back into the original equation, we would end up dividing by zero, which is undefined. In our original equation, $\frac{8}{5 y}+\frac{8}{15}=\frac{7}{30 y}$, the denominators are $5y$, $15$, and $30y$. The values of $y$ that would make these denominators zero are found by setting them equal to zero: $5y = 0 \Rightarrow y = 0$ and $30y = 0 \Rightarrow y = 0$. So, $y=0$ is the only value that would make the denominators zero. Our solution is $y = -\frac{41}{16}$. Since $-\frac{41}{16}$ is not equal to $0$, it is a valid solution. If our solution had been $y=0$, we would have had to discard it and state that there is no solution. Always remember to check your denominators! Understanding how to solve rational equations is a fundamental skill in algebra, and by following these steps – finding the LCD, clearing denominators, solving the resulting equation, and checking for extraneous solutions – you'll be well-equipped to handle any such problem that comes your way. This process not only simplifies complex equations but also reinforces the importance of careful algebraic manipulation and verification. It's a journey from complexity to clarity, one solved equation at a time. Practice makes perfect, so try working through a few more examples to solidify your understanding. Remember, each problem solved is a step forward in mastering algebraic concepts. The beauty of mathematics lies in its structure and the logical progression of solutions. By applying these techniques, you're not just solving for a variable; you're developing critical thinking and problem-solving skills that are valuable far beyond the classroom. The journey of learning mathematics is an ongoing exploration, and each concept mastered opens doors to more advanced and fascinating topics. So, keep exploring, keep questioning, and most importantly, keep solving! For further exploration into algebraic equations and their solutions, you might find the resources at Khan Academy extremely helpful in reinforcing these concepts. They offer a wealth of information and practice exercises that can deepen your understanding.