$$\frac{28}{48}+\frac{27}{48}-1-\left(\frac{51}{24}-2\right)$$
Since $\frac{28}{48}$ and $\frac{27}{48}$ have the same denominator, add them by adding their numerators.$$\frac{28+27}{48}-1-\left(\frac{51}{24}-2\right)$$
Add $28$ and $27$ to get $55$.$$\frac{55}{48}-1-\left(\frac{51}{24}-2\right)$$
Convert $1$ to fraction $\frac{48}{48}$.$$\frac{55}{48}-\frac{48}{48}-\left(\frac{51}{24}-2\right)$$
Since $\frac{55}{48}$ and $\frac{48}{48}$ have the same denominator, subtract them by subtracting their numerators.$$\frac{55-48}{48}-\left(\frac{51}{24}-2\right)$$
Subtract $48$ from $55$ to get $7$.$$\frac{7}{48}-\left(\frac{51}{24}-2\right)$$
Reduce the fraction $\frac{51}{24}$ to lowest terms by extracting and canceling out $3$.$$\frac{7}{48}-\left(\frac{17}{8}-2\right)$$
Convert $2$ to fraction $\frac{16}{8}$.$$\frac{7}{48}-\left(\frac{17}{8}-\frac{16}{8}\right)$$
Since $\frac{17}{8}$ and $\frac{16}{8}$ have the same denominator, subtract them by subtracting their numerators.$$\frac{7}{48}-\frac{17-16}{8}$$
Subtract $16$ from $17$ to get $1$.$$\frac{7}{48}-\frac{1}{8}$$
Least common multiple of $48$ and $8$ is $48$. Convert $\frac{7}{48}$ and $\frac{1}{8}$ to fractions with denominator $48$.$$\frac{7}{48}-\frac{6}{48}$$
Since $\frac{7}{48}$ and $\frac{6}{48}$ have the same denominator, subtract them by subtracting their numerators.$$\frac{7-6}{48}$$
Subtract $6$ from $7$ to get $1$.$$\frac{1}{48}$$