Solving Rational Equations: Find The Value Of T

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Are you wrestling with rational equations? Do you find yourself staring at fractions with variables and wondering where to even begin? Well, you're not alone! Rational equations can seem daunting at first, but with a step-by-step approach and a little algebraic know-how, you can conquer them. In this guide, we'll break down the process of solving a specific rational equation and provide you with the tools you need to tackle similar problems.

Understanding Rational Equations

Before we dive into the solution, let's define what a rational equation is. In essence, a rational equation is an equation that contains one or more fractions where the numerator and/or denominator are polynomials. Our example equation, $\frac{t+7}{t-2}=\frac{5}{t}+3$, perfectly fits this description. Notice the fractions with 't' in both the numerator and denominator. The key to solving these equations lies in eliminating the fractions, which we'll do by finding a common denominator.

When we talk about solving equations, what we are really trying to do is find what value(s) of the variable(s) will make the equation a true statement. We can think of an equation like a balanced scale: whatever operation we perform on one side, we must perform on the other to maintain the balance. With rational equations, we need to be particularly cautious about values that would make the denominator zero, as division by zero is undefined. These are called extraneous solutions, and we'll keep an eye out for them as we solve our equation.

The goal here is not just to get the right answer but to understand the process. Mathematics is not merely a collection of formulas, but a way of thinking. By carefully following each step and understanding the reasoning behind it, you will not only be able to solve this specific problem but also a wide range of similar problems. Think of it as learning a skill, like cooking or playing a musical instrument; the more you practice and understand the fundamentals, the better you become.

The Equation at Hand

Let's revisit the equation we're going to solve:

t+7t−2=5t+3\frac{t+7}{t-2}=\frac{5}{t}+3

Our mission is to find the value(s) of 't' that make this equation true. The options provided are:

A. $t=0$ B. $t=-1$ C. $t=2$ D. $t=-3$ E. $t=5$

We could try plugging in each of these values to see which ones work, but that can be time-consuming. Instead, we'll use a systematic algebraic approach to find the solution(s). This method will not only give us the answer but also deepen our understanding of rational equations.

Step-by-Step Solution

Step 1: Identify Restrictions

Before we start manipulating the equation, it's crucial to identify any values of 't' that would make the denominators zero. These values are not allowed because division by zero is undefined. Looking at our equation, we have two denominators: t-2 and t. So:

  • t-2 ≠ 0 implies t ≠ 2
  • t ≠ 0

This means that t cannot be 0 or 2. These are our restrictions, and if we get these values as solutions later, we'll need to discard them.

Step 2: Find the Least Common Denominator (LCD)

To eliminate the fractions, we need to multiply both sides of the equation by the least common denominator (LCD). The LCD is the smallest expression that is divisible by all the denominators in the equation. In our case, the denominators are t-2 and t. Since they don't share any common factors, the LCD is simply their product: t(t-2). Finding the LCD is an important step because it ensures that we can clear all the fractions in one go.

Step 3: Multiply Both Sides by the LCD

Now we'll multiply both sides of the equation by the LCD, t(t-2):

t(t−2)⋅t+7t−2=t(t−2)⋅(5t+3)t(t-2) \cdot \frac{t+7}{t-2} = t(t-2) \cdot \left(\frac{5}{t} + 3\right)

This step is the heart of the solution process. By multiplying both sides by the LCD, we eliminate the denominators, which transforms our rational equation into a more manageable polynomial equation.

Step 4: Simplify

Next, we simplify both sides of the equation. On the left side, the (t-2) terms cancel out:

t(t+7)=t(t−2)⋅(5t+3)t(t+7) = t(t-2) \cdot \left(\frac{5}{t} + 3\right)

On the right side, we need to distribute t(t-2) to both terms inside the parentheses:

t(t+7)=t(t−2)⋅5t+t(t−2)⋅3t(t+7) = t(t-2) \cdot \frac{5}{t} + t(t-2) \cdot 3

Now we can simplify further. The t terms cancel in the first part of the right side:

t(t+7)=(t−2)⋅5+t(t−2)⋅3t(t+7) = (t-2) \cdot 5 + t(t-2) \cdot 3

This simplification makes the equation much easier to work with. We've successfully cleared the fractions, and now we have a polynomial equation to solve.

Step 5: Distribute and Expand

Let's distribute and expand the terms on both sides of the equation:

t2+7t=5t−10+3t2−6tt^2 + 7t = 5t - 10 + 3t^2 - 6t

This step involves basic algebraic manipulation, but it's crucial for setting up the equation to be solved. Remember to distribute carefully and pay attention to signs.

Step 6: Combine Like Terms

Now we combine like terms on the right side of the equation:

t2+7t=3t2−t−10t^2 + 7t = 3t^2 - t - 10

Combining like terms simplifies the equation and makes it easier to see the structure of the polynomial.

Step 7: Set the Equation to Zero

To solve the quadratic equation, we need to set it equal to zero. Subtract t^2 and 7t from both sides:

0=2t2−8t−100 = 2t^2 - 8t - 10

Setting the equation to zero allows us to use factoring or the quadratic formula to find the solutions.

Step 8: Simplify the Quadratic Equation

Notice that all the coefficients are even, so we can divide the entire equation by 2 to simplify it:

0=t2−4t−50 = t^2 - 4t - 5

Simplifying the equation makes it easier to factor or apply the quadratic formula. This step can save you time and reduce the chance of errors.

Step 9: Factor the Quadratic Equation

Now we factor the quadratic equation:

0=(t−5)(t+1)0 = (t - 5)(t + 1)

Factoring is a powerful technique for solving quadratic equations. It allows us to find the values of t that make the equation true by setting each factor equal to zero.

Step 10: Solve for t

Set each factor equal to zero and solve for t:

  • t - 5 = 0 implies t = 5
  • t + 1 = 0 implies t = -1

So we have two potential solutions: t = 5 and t = -1.

Step 11: Check for Extraneous Solutions

Remember our restrictions from Step 1? We found that t cannot be 0 or 2. Our solutions, 5 and -1, are not in this restricted set, so they are both valid.

Step 12: Final Answer

The values of t that satisfy the equation are 5 and -1. Therefore, the correct answers are:

  • E. $t=5$
  • B. $t=-1$

Conclusion

Solving rational equations involves a series of algebraic steps, from identifying restrictions to factoring quadratic equations. By following this step-by-step guide, you can confidently tackle similar problems. Remember to always check for extraneous solutions to ensure the validity of your answers. Keep practicing, and you'll become a rational equation-solving pro in no time!

For more in-depth explanations and examples, you can explore resources like Khan Academy's Algebra I section. This is an excellent resource for reinforcing your understanding of rational equations and other algebraic concepts.