11. \( 2\left(3p+\frac{1}{4}\right)=5\left(2p-\frac{1}{6}\right) \)

To solve the equation \( 2\left(3p+\frac{1}{4}\right)=5\left(2p-\frac{1}{6}\right) \), we start by distributing both sides. On the left side, we have: \[ 2 \cdot 3p + 2 \cdot \frac{1}{4} = 6p + \frac{1}{2} \] On the right side, we have: \[ 5 \cdot 2p - 5 \cdot \frac{1}{6} = 10p - \frac{5}{6} \] Now, our equation looks like this: \[ 6p + \frac{1}{2} = 10p - \frac{5}{6} \] Next, we want to eliminate the fractions. The least common multiple of 2 and 6 is 6, so we multiply every term by 6: \[ 6(6p) + 6\left(\frac{1}{2}\right) = 6(10p) - 6\left(\frac{5}{6}\right) \] This simplifies to: \[ 36p + 3 = 60p - 5 \] Now, let's move all terms involving \( p \) to one side and constant terms to the other side: \[ 36p - 60p = -5 - 3 \] which simplifies to: \[ -24p = -8 \] Finally, we divide both sides by -24: \[ p = \frac{-8}{-24} = \frac{1}{3} \] Thus, the solution is: \[ p = \frac{1}{3} \]