Problem 31 - College Algebra with Modeling and Visualization

Question

Solve the equation and check your answer. $$ \frac{1}{2}(d-3)-\frac{2}{3}(2 d-5)=\frac{5}{12} $$

Step-by-Step Solution

Answer

The solution is \( d = \frac{17}{10} \).

Step 1: Distribute the Fraction Coefficients

To eliminate the parentheses, distribute the fractions over the terms inside the parentheses. For \( \frac{1}{2}(d-3) \), distribute the \( \frac{1}{2} \): \[ \frac{1}{2}(d-3) = \frac{1}{2}d - \frac{1}{2} imes 3 = \frac{1}{2}d - \frac{3}{2} \]For \( -\frac{2}{3}(2 d-5) \), distribute the \( -\frac{2}{3} \):\[-\frac{2}{3}(2d-5) = -\frac{2}{3} imes 2d + \frac{2}{3} imes 5 = -\frac{4}{3}d + \frac{10}{3} \]Now, substitute back into the equation:\[ \frac{1}{2}d - \frac{3}{2} - \frac{4}{3}d + \frac{10}{3} = \frac{5}{12} \]

Step 2: Combine Like Terms

Combine terms involving \( d \) and numerical terms separately.Combine the \( d \) terms: \[ \frac{1}{2}d - \frac{4}{3}d = \left( \frac{3}{6} - \frac{8}{6} \right)d = -\frac{5}{6}d \]Combine the constant terms:\[- \frac{3}{2} + \frac{10}{3} = -\frac{9}{6} + \frac{20}{6} = \frac{11}{6} \]The equation simplifies to:\[ -\frac{5}{6}d + \frac{11}{6} = \frac{5}{12} \]

Step 3: Eliminate Fractions by Multiplying

To eliminate fractions, multiply the entire equation by the least common multiple (LCM) of the denominators. Here, the LCM of 6 and 12 is 12.Multiply through by 12:\[ 12\left(-\frac{5}{6}d + \frac{11}{6}\right) = 12\left(\frac{5}{12}\right) \]Simplifying each term:\[ -10d + 22 = 5 \]

Step 4: Isolate d by Solving the Equation

Subtract 22 from both sides to isolate the d term:\[ -10d = 5 - 22 = -17 \]Divide both sides by -10 to solve for \( d \):\[ d = \frac{-17}{-10} = \frac{17}{10} \]

Step 5: Check the Solution

Substitute \( \frac{17}{10} \) back into the original equation to verify.For \( \frac{1}{2}(d - 3) \):\[ \frac{1}{2}\left(\frac{17}{10} - 3\right) = \frac{1}{2} \times \left(\frac{17}{10} - \frac{30}{10}\right) = \frac{1}{2} \times \left(-\frac{13}{10}\right) = -\frac{13}{20} \]For \( -\frac{2}{3}(2d - 5) \):\[ -\frac{2}{3} \left(2 \times \frac{17}{10} - 5 \right) = -\frac{2}{3}\left(\frac{34}{10} - \frac{50}{10}\right)= -\frac{2}{3} \times \left(-\frac{16}{10}\right) = \frac{32}{30} = \frac{16}{15} \]Combine to check:\[-\frac{13}{20} + \frac{16}{15} = \frac{-39}{60} + \frac{64}{60} = \frac{25}{60} = \frac{5}{12} \]The solution \( d = \frac{17}{10} \) is correct.

Key Concepts

Fraction Distribution

When solving equations with fractions, one useful tool is to distribute the fraction across terms inside parentheses. This technique breaks down expressions into simpler parts for easy handling.
For example, if we start with an expression like \( \frac{1}{2}(d-3) \), the act of distribution involves multiplying each term inside the parentheses by \( \frac{1}{2} \).
This gives us \( \frac{1}{2} \times d \) and \( \frac{1}{2} \times -3 \), which simplifies to \( \frac{1}{2}d - \frac{3}{2} \).
The same approach is used for \( -\frac{2}{3}(2d-5) \). Distributing \( -\frac{2}{3} \) through the terms yields \( -\frac{2}{3} \times 2d \) and \( -\frac{2}{3} \times -5 \).
This simplifies further to \( -\frac{4}{3}d + \frac{10}{3} \). By distributing fractions, we make each component of the equation easier to manage, paving the way for combining like terms.

Combining Like Terms

After distributing fractions, the next essential step in equation solving is combining like terms. Like terms have the same variables raised to the same powers and are combined by simply adding or subtracting constants.
In our equation: \( \frac{1}{2}d - \frac{4}{3}d + \frac{11}{6} = \frac{5}{12} \), we focus first on the terms involving \( d \).
To combine \( \frac{1}{2}d \) and \( -\frac{4}{3}d \), we convert these fractions to have a common denominator.
This involves rewriting them as \( \frac{3}{6}d \) and \( -\frac{8}{6}d \) so that we can directly combine them to get \( -\frac{5}{6}d \).
Next, we combine the constant terms, \( -\frac{3}{2} \) and \( \frac{10}{3} \), by rewriting them with a common denominator of 6, resulting in \( -\frac{9}{6} + \frac{20}{6} \), which simplifies to \( \frac{11}{6} \).

• This simplification refines our equations, making the solution process converge more readily.

LCM in Equations

Eliminating fractions simplifies equations and is achieved by using the Least Common Multiple (LCM) of the denominators. This step is vital to clearing fractions for straightforward arithmetic.
In this exercise, we found that the LCM of our denominators, 6 and 12, is 12. By multiplying the entire equation by this LCM, each term's fractions disappear.
This means taking the equation \( -\frac{5}{6}d + \frac{11}{6} = \frac{5}{12} \) and multiplying every term by 12. This simplifies to \( -10d + 22 = 5 \).

• By clearing fractions, all parts of the equation become whole numbers, making algebraic manipulation more intuitive and manageable.

Verification of Solution

Once a solution is reached, it's essential to verify it by substituting the solution back into the original equation. This checks for any calculation errors and ensures authenticity. In our case, the candidate solution is \( d = \frac{17}{10} \).
We substituted \( \frac{17}{10} \) back into the individual expressions: \( \frac{1}{2}(d - 3) \) and \( -\frac{2}{3}(2d - 5) \).
Each substitution followed by recalculation resulted in \( -\frac{13}{20} \) and \( \frac{16}{15} \) respectively, ultimately verifying the left side computes to the original right-hand side of \( \frac{5}{12} \).

• Verification acts as a safety check to confirm the correctness and integrity of the solution obtained from the equation.