12(х+1 1/3)=5(21/5х+7 4/15)

To solve the equation \( 12 \left( x + 1 \frac{1}{3} \right) = 5 \left( \frac{21}{5} x + 7 \frac{4}{15} \right) \), we will first convert the mixed numbers to improper fractions.

1. Convert \( 1 \frac{1}{3} \) to an improper fraction: \[ 1 \frac{1}{3} = \frac{3 \times 1 + 1}{3} = \frac{4}{3} \]

2. Convert \( 7 \frac{4}{15} \) to an improper fraction: \[ 7 \frac{4}{15} = \frac{7 \times 15 + 4}{15} = \frac{105 + 4}{15} = \frac{109}{15} \]

Now, substitute these values back into the equation:

\[ 12 \left( x + \frac{4}{3} \right) = 5 \left( \frac{21}{5} x + \frac{109}{15} \right) \]

This simplifies to:

\[ 12x + 12 \cdot \frac{4}{3} = 5 \cdot \frac{21}{5} x + 5 \cdot \frac{109}{15} \]

Calculating \( 12 \cdot \frac{4}{3} \):

\[ 12 \cdot \frac{4}{3} = 4 \cdot 4 = 16 \]

Calculating \( 5 \cdot \frac{21}{5} x \): \[ 5 \cdot \frac{21}{5} x = 21x \]

Calculating \( 5 \cdot \frac{109}{15} \):

\[ 5 \cdot \frac{109}{15} = \frac{545}{15} \]

Now the equation looks like:

\[ 12x + 16 = 21x + \frac{545}{15} \]

Next, we can eliminate the fractions by multiplying the entire equation by 15:

\[ 15(12x + 16) = 15 \left( 21x + \frac{545}{15} \right) \]

Which simplifies to:

\[ 180x + 240 = 315x + 545 \]

Now, let's move all \( x \) terms to one side and constant terms to the other:

\[ 180x - 315x = 545 - 240 \]

This simplifies to:

\[ -135x = 305 \]

Now, divide by -135:

\[ x = \frac{305}{-135} = -\frac{61}{27} \]

Thus, the solution to the equation is:

\[ x = -\frac{61}{27} \]