Solving U/(4u-3) = 2/(5u-9) A Step-by-Step Guide

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In this article, we will delve into the step-by-step process of solving the equation u4u−3=25u−9\frac{u}{4u-3} = \frac{2}{5u-9}. This equation involves fractions with algebraic expressions in the denominator, requiring careful manipulation and application of algebraic principles to arrive at the solution. Our goal is to find the values of 'u' that satisfy the given equation, expressing each answer as either an integer or a fraction in its simplest form. This exploration is crucial for students and enthusiasts looking to enhance their skills in solving rational equations, a fundamental topic in algebra.

Understanding Rational Equations

Rational equations, like the one we're tackling, involve fractions where the numerator and/or the denominator contain variables. The key to solving these equations lies in eliminating the fractions, which can be achieved by multiplying both sides of the equation by the least common denominator (LCD). However, it's absolutely crucial to identify any values of the variable that would make the denominator zero, as these values are excluded from the solution set. These values are known as extraneous solutions and must be checked at the end of the solution process.

Step 1: Identifying Potential Extraneous Solutions

Before we begin manipulating the equation, let's first identify any values of 'u' that would make the denominators zero. This is essential for spotting extraneous solutions later on. We have two denominators in our equation: (4u−3)(4u - 3) and (5u−9)(5u - 9). Setting each of these equal to zero, we get:

  • 4u−3=04u - 3 = 0 => 4u=34u = 3 => u=34u = \frac{3}{4}
  • 5u−9=05u - 9 = 0 => 5u=95u = 9 => u=95u = \frac{9}{5}

Therefore, u=34u = \frac{3}{4} and u=95u = \frac{9}{5} are potential extraneous solutions and must be checked at the end.

Step 2: Eliminating the Fractions

To eliminate the fractions, we need to multiply both sides of the equation by the least common denominator (LCD). In this case, the LCD is the product of the two denominators, which is (4u−3)(5u−9)(4u - 3)(5u - 9). Multiplying both sides of the equation by the LCD, we get:

u4u−3⋅(4u−3)(5u−9)=25u−9⋅(4u−3)(5u−9)\frac{u}{4u-3} \cdot (4u-3)(5u-9) = \frac{2}{5u-9} \cdot (4u-3)(5u-9)

This simplifies to:

u(5u−9)=2(4u−3)u(5u - 9) = 2(4u - 3)

This step effectively clears the fractions, leaving us with a quadratic equation to solve.

Step 3: Simplifying and Rearranging the Equation

Now, let's expand both sides of the equation:

5u2−9u=8u−65u^2 - 9u = 8u - 6

Next, we want to rearrange the equation into the standard quadratic form, which is ax2+bx+c=0ax^2 + bx + c = 0. To do this, we subtract 8u8u and add 66 to both sides:

5u2−9u−8u+6=05u^2 - 9u - 8u + 6 = 0

5u2−17u+6=05u^2 - 17u + 6 = 0

We now have a quadratic equation in standard form.

Step 4: Solving the Quadratic Equation

To solve the quadratic equation 5u2−17u+6=05u^2 - 17u + 6 = 0, we can use several methods, such as factoring, completing the square, or using the quadratic formula. In this case, factoring seems to be the most straightforward approach. We are looking for two numbers that multiply to (5)(6)=30(5)(6) = 30 and add up to −17-17. These numbers are −15-15 and −2-2. So we can rewrite the middle term as:

5u2−15u−2u+6=05u^2 - 15u - 2u + 6 = 0

Now, we can factor by grouping:

5u(u−3)−2(u−3)=05u(u - 3) - 2(u - 3) = 0

(5u−2)(u−3)=0(5u - 2)(u - 3) = 0

Setting each factor equal to zero, we get:

  • 5u−2=05u - 2 = 0 => 5u=25u = 2 => u=25u = \frac{2}{5}
  • u−3=0u - 3 = 0 => u=3u = 3

Thus, we have two potential solutions: u=25u = \frac{2}{5} and u=3u = 3.

Step 5: Checking for Extraneous Solutions

It's absolutely vital that we check our potential solutions against the values we identified earlier as potential extraneous solutions (u=34u = \frac{3}{4} and u=95u = \frac{9}{5}). Neither of our solutions, u=25u = \frac{2}{5} and u=3u = 3, match these extraneous values. Therefore, we need to substitute our solutions back into the original equation to ensure they satisfy it.

Checking u=25u = \frac{2}{5}:

254(25)−3=25(25)−9\frac{\frac{2}{5}}{4(\frac{2}{5})-3} = \frac{2}{5(\frac{2}{5})-9}

2585−3=22−9\frac{\frac{2}{5}}{\frac{8}{5}-3} = \frac{2}{2-9}

258−155=2−7\frac{\frac{2}{5}}{\frac{8-15}{5}} = \frac{2}{-7}

25−75=2−7\frac{\frac{2}{5}}{\frac{-7}{5}} = \frac{2}{-7}

25⋅5−7=2−7\frac{2}{5} \cdot \frac{5}{-7} = \frac{2}{-7}

−27=−27\frac{-2}{7} = \frac{-2}{7}

This solution checks out.

Checking u=3u = 3:

34(3)−3=25(3)−9\frac{3}{4(3)-3} = \frac{2}{5(3)-9}

312−3=215−9\frac{3}{12-3} = \frac{2}{15-9}

39=26\frac{3}{9} = \frac{2}{6}

13=13\frac{1}{3} = \frac{1}{3}

This solution also checks out.

Step 6: Stating the Solutions

Since both potential solutions satisfy the original equation and are not extraneous, we can confidently state that the solutions are:

u=25 and u=3u = \frac{2}{5} \text{ and } u = 3

These are the values of 'u' that make the equation u4u−3=25u−9\frac{u}{4u-3} = \frac{2}{5u-9} true.

Conclusion

In this comprehensive guide, we have meticulously walked through the process of solving the rational equation u4u−3=25u−9\frac{u}{4u-3} = \frac{2}{5u-9}. We emphasized the crucial importance of identifying potential extraneous solutions before manipulating the equation, a step that often gets overlooked but is critical for accuracy. We then systematically eliminated the fractions, simplified the resulting quadratic equation, and solved it by factoring. Finally, we rigorously checked our solutions to ensure they were valid. This step-by-step approach not only provides the correct answers but also reinforces the fundamental principles of solving rational equations. Remember, practice makes perfect, and consistently applying these principles will solidify your understanding and boost your confidence in tackling similar algebraic challenges.

Common Mistakes to Avoid When Solving Rational Equations

Solving rational equations can be tricky, and it's easy to make mistakes if you're not careful. Here are some common pitfalls to avoid:

  • Forgetting to Identify Extraneous Solutions: This is arguably the most common mistake. Always check for values that make the denominators zero before you start solving. Failing to do so can lead you to include extraneous solutions in your final answer.
  • Incorrectly Applying the Distributive Property: When multiplying both sides of the equation by the LCD, make sure you distribute correctly to every term. A missed term can throw off the entire solution.
  • Making Sign Errors: Sign errors are easy to make when rearranging equations or factoring. Double-check your work, especially when dealing with negative signs.
  • Incorrectly Factoring Quadratic Equations: If you choose to solve a quadratic equation by factoring, ensure you factor correctly. If you're unsure, use the quadratic formula instead.
  • Not Checking Your Solutions: Even if you've been careful, it's always a good idea to plug your solutions back into the original equation to verify they work. This is especially important for rational equations.

By being aware of these common mistakes, you can increase your chances of solving rational equations accurately and efficiently.

Practice Problems

To further solidify your understanding, try solving these practice problems:

  1. x2x+1=1x−2\frac{x}{2x+1} = \frac{1}{x-2}
  2. 3y−1+2y+1=5y2−1\frac{3}{y-1} + \frac{2}{y+1} = \frac{5}{y^2-1}
  3. a+2a−3=a−1a+2\frac{a+2}{a-3} = \frac{a-1}{a+2}

By working through these problems, you'll gain valuable experience and refine your problem-solving skills.

Further Exploration

If you're interested in learning more about rational equations, consider exploring these topics:

  • Solving Rational Inequalities: This extends the concepts of solving rational equations to inequalities.
  • Applications of Rational Equations: Rational equations have applications in various fields, such as physics, engineering, and economics.
  • More Complex Rational Expressions: Practice simplifying and manipulating more complex rational expressions to build your algebraic fluency.

By delving deeper into these areas, you'll develop a more comprehensive understanding of rational equations and their applications.

The original question was: "Solve u4u−3=25u−9\frac{u}{4 u-3}=\frac{2}{5 u-9}. Give each answer as an integer or as a fraction in its simplest form." A clearer and more easily understood version of the question is: "Find the solutions for the equation u4u−3=25u−9\frac{u}{4u-3} = \frac{2}{5u-9}, expressing each solution as an integer or a simplified fraction."

Solving Rational Equations A Step-by-Step Guide to u/(4u-3) = 2/(5u-9)