Question

$\frac{3}{5}\times\left(\frac{1}{3}+\frac{1}{8}\
ight)=\\____________\\
\left(\frac{1}{2}+\frac{2}{3}\
ight)\times\left(\frac{5}{6}-\frac{1}{4}\
ight)=\\____________$

Explanation:

Step1: Solve the first equation

First, calculate the sum inside the parentheses: $\frac{1}{3} + \frac{1}{8}$. Find a common denominator, which is 24. So $\frac{1}{3} = \frac{8}{24}$ and $\frac{1}{8} = \frac{3}{24}$. Then $\frac{8}{24} + \frac{3}{24} = \frac{11}{24}$. Now multiply by $\frac{3}{5}$: $\frac{3}{5} \times \frac{11}{24} = \frac{33}{120} = \frac{11}{40}$.

Step2: Solve the second equation

First, calculate the sum inside the first parentheses: $\frac{1}{2} + \frac{2}{3}$. The common denominator is 6. So $\frac{1}{2} = \frac{3}{6}$ and $\frac{2}{3} = \frac{4}{6}$. Then $\frac{3}{6} + \frac{4}{6} = \frac{7}{6}$. Next, calculate the difference inside the second parentheses: $\frac{5}{6} - \frac{1}{4}$. The common denominator is 12. So $\frac{5}{6} = \frac{10}{12}$ and $\frac{1}{4} = \frac{3}{12}$. Then $\frac{10}{12} - \frac{3}{12} = \frac{7}{12}$. Now multiply the two results: $\frac{7}{6} \times \frac{7}{12} = \frac{49}{72}$.

Answer:

First equation: $\frac{11}{40}$; Second equation: $\frac{49}{72}$