Verified Answer :

  • Convert all mixed numbers to improper fractions: \(1 \frac{2}{5} = \frac{7}{5}\), \(3 \frac{1}{2} = \frac{7}{2}\), and \(5 \frac{5}{6} = \frac{35}{6}\).
  • Rewrite the equation as \(\frac{7}{5} k - \frac{7}{2} = -\frac{35}{6}\).
  • Add \(\frac{7}{2}\) to both sides to isolate the \(k\) term: \(\frac{7}{5} k = -\frac{14}{6} = -\frac{7}{3}\).
  • Multiply by the reciprocal \(\frac{5}{7}\) to find the solution: \(k = \boxed{-\frac{5}{3}}\).

Explanation

  1. Problem Analysis and Strategy

To solve for the variable \(k\) in the equation \(1 \frac{2}{5} k - 3 \frac{1}{2} = -5 \frac{5}{6}\), we first analyze the structure of the equation. It is a linear equation in the form \(Ak + B = C\), where \(A, B, C\) are mixed numbers. Our strategy is to convert these mixed numbers into improper fractions to simplify the algebraic manipulation, isolate \(k\), and solve for its exact value.

  1. Converting to Improper Fractions

First, we convert the mixed fractions into improper fractions:

  • \(1 \frac{2}{5} = \frac{1 \times 5 + 2}{5} = \frac{7}{5}\)
  • \(3 \frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{7}{2}\)
  • \(5 \frac{5}{6} = \frac{5 \times 6 + 5}{6} = \frac{35}{6}\)

Substituting these values into the original equation, we obtain:
\(\frac{7}{5} k - \frac{7}{2} = -\frac{35}{6}\)

  1. Isolating the Term with k

To isolate the term with \(k\), we add \(\frac{7}{2}\) to both sides of the equation:
\(\frac{7}{5} k = -\frac{35}{6} + \frac{7}{2}\)

To add the fractions on the right-hand side, we need a common denominator, which is \(6\):
\(\frac{7}{2} = \frac{7 \times 3}{2 \times 3} = \frac{21}{6}\)

Now, perform the addition:
\(\frac{7}{5} k = -\frac{35}{6} + \frac{21}{6} = \frac{-35 + 21}{6} = -\frac{14}{6}\)
Reducing the fraction \(-\frac{14}{6}\) by dividing both numerator and denominator by \(2\), we get:
\(\frac{7}{5} k = -\frac{7}{3}\)

  1. Solving for k

To solve for \(k\), we multiply both sides of the equation by the reciprocal of \(\frac{7}{5}\), which is \(\frac{5}{7}\):
\(k = -\frac{7}{3} \times \frac{5}{7}\)

The \(7\)s in the numerator and denominator cancel out:
\(k = -\frac{1 \times 5}{3 \times 1} = -\frac{5}{3}\)

Thus, the final value of \(k\) is \(-\frac{5}{3}\) (or \(-1 \frac{2}{3}\)).

Examples

Understanding how to solve linear equations with fractions is essential for managing real-world quantities, such as calculating changes in inventory or financial balances. For instance, if a business starts with a certain amount of stock expressed as a mixed fraction and undergoes changes described by linear operations, solving for the final quantity requires these exact algebraic techniques. Mastering this ensures accurate resource management and helps in predicting outcomes in any situation involving rates or proportional changes.

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