Solving For T In The Equation (2/3)t - (1/5)t = 2

by Admin 50 views

At the heart of this mathematical problem lies the task of solving for t in the given equation: 23t−15t=2\frac{2}{3} t - \frac{1}{5} t = 2. This is a fundamental algebraic problem that requires us to isolate the variable t and determine its value. Understanding how to solve such equations is crucial for various applications in mathematics, science, and engineering. Let's delve into the step-by-step process of finding the correct solution.

Before we dive into the solution, it's important to grasp the basic principles involved. We are dealing with a linear equation in one variable. The goal is to manipulate the equation using algebraic operations while maintaining equality, until we have t isolated on one side. This involves combining like terms, performing arithmetic operations, and applying the properties of equality. The ability to confidently tackle such problems forms a cornerstone of mathematical proficiency.

To successfully navigate this problem, we need to recall the order of operations and the principles of fraction manipulation. We'll be working with fractions, so finding a common denominator will be essential for combining the terms involving t. We'll also need to remember that whatever operation we perform on one side of the equation, we must perform on the other side to maintain balance. This ensures that the equality remains valid throughout the solution process. Let's begin by addressing the fractional coefficients of t and simplifying the equation.

  1. Combine the t terms: To begin, we need to combine the terms involving t. This means finding a common denominator for the fractions 23\frac{2}{3} and 15\frac{1}{5}. The least common multiple of 3 and 5 is 15. So, we rewrite the fractions with the common denominator:

    23t−15t=2×53×5t−1×35×3t=1015t−315t\frac{2}{3} t - \frac{1}{5} t = \frac{2 \times 5}{3 \times 5} t - \frac{1 \times 3}{5 \times 3} t = \frac{10}{15} t - \frac{3}{15} t

    Now, we can subtract the fractions:

    1015t−315t=10−315t=715t\frac{10}{15} t - \frac{3}{15} t = \frac{10-3}{15} t = \frac{7}{15} t

    So, our equation now looks like this:

    715t=2\frac{7}{15} t = 2

  2. Isolate t: To isolate t, we need to get rid of the fraction 715\frac{7}{15} that is multiplying it. We can do this by multiplying both sides of the equation by the reciprocal of 715\frac{7}{15}, which is 157\frac{15}{7}:

    157×715t=2×157\frac{15}{7} \times \frac{7}{15} t = 2 \times \frac{15}{7}

    On the left side, the fractions cancel out, leaving us with just t:

    t=2×157t = 2 \times \frac{15}{7}

  3. Calculate the value of t: Now we simply multiply the right side:

    t=2×157=307t = \frac{2 \times 15}{7} = \frac{30}{7}

    Therefore, the solution for t is:

    t=307t = \frac{30}{7}

Now that we have found the solution, let's compare it with the given answer choices:

  • A. t = 6
  • B. t = 307\frac{30}{7}
  • C. t = 730\frac{7}{30}
  • D. t = 23\frac{2}{3}

Our calculated solution, t = 307\frac{30}{7}, matches answer choice B. Therefore, the correct answer is B.

It's always a good practice to check our solution by plugging it back into the original equation. Let's substitute t = 307\frac{30}{7} into the equation 23t−15t=2\frac{2}{3} t - \frac{1}{5} t = 2:

23(307)−15(307)=2×303×7−1×305×7=6021−3035\frac{2}{3} \left( \frac{30}{7} \right) - \frac{1}{5} \left( \frac{30}{7} \right) = \frac{2 \times 30}{3 \times 7} - \frac{1 \times 30}{5 \times 7} = \frac{60}{21} - \frac{30}{35}

To subtract these fractions, we need a common denominator. The least common multiple of 21 and 35 is 105. So, we rewrite the fractions:

6021−3035=60×521×5−30×335×3=300105−90105=300−90105=210105=2\frac{60}{21} - \frac{30}{35} = \frac{60 \times 5}{21 \times 5} - \frac{30 \times 3}{35 \times 3} = \frac{300}{105} - \frac{90}{105} = \frac{300 - 90}{105} = \frac{210}{105} = 2

The left side of the equation equals 2, which matches the right side. This confirms that our solution, t = 307\frac{30}{7}, is correct.

When solving equations like this, several common mistakes can lead to incorrect answers. Understanding these pitfalls can help you avoid them and improve your problem-solving accuracy.

  1. Incorrectly Combining Fractions: One frequent error is making mistakes while finding a common denominator or subtracting the fractions. For example, students might incorrectly add or subtract the numerators or denominators without finding a common denominator first. To avoid this, always double-check your arithmetic and ensure that you have correctly found the least common multiple and adjusted the numerators accordingly.
  2. Forgetting to Multiply Both Sides: Another common mistake is forgetting to perform the same operation on both sides of the equation. For instance, when multiplying by the reciprocal to isolate t, students might only multiply one side, thus disrupting the equality. Always remember that any operation performed on one side must be mirrored on the other side to maintain the balance of the equation.
  3. Arithmetic Errors: Simple arithmetic errors, such as miscalculations in multiplication or division, can easily lead to incorrect results. It's crucial to be meticulous in your calculations and double-check your work. Using a calculator for complex arithmetic can also help reduce the likelihood of errors.
  4. Misunderstanding the Order of Operations: Students may sometimes misapply the order of operations (PEMDAS/BODMAS), leading to incorrect simplification. Ensure that you are performing operations in the correct order: Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

By being aware of these common mistakes and taking steps to avoid them, you can significantly improve your accuracy in solving algebraic equations.

In summary, the solution for t in the equation 23t−15t=2\frac{2}{3} t - \frac{1}{5} t = 2 is B. t = 307\frac{30}{7}. This problem demonstrates the fundamental principles of solving linear equations, including combining like terms, working with fractions, and isolating the variable. By understanding these steps and avoiding common mistakes, you can confidently tackle similar problems in mathematics and beyond. Remember to always check your solution by plugging it back into the original equation to ensure accuracy.