\(\displaystyle \frac{1}{4}+\frac{1}{2}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{1}{2}\times\frac{2}{2}=\frac{2}{4}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{1}{4}+\frac{2}{4}=\frac{3}{4}\)

\(\displaystyle \frac{1}{5}+\frac{1}{2}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{1}{5}\times\frac{2}{2}=\frac{2}{10}\)

\(\displaystyle \frac{1}{2}\times \frac{5}{5}=\frac{5}{10}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{2}{10}+\frac{5}{10}=\frac{7}{10}\)

\(\displaystyle \frac{1}{3}+\frac{2}{5}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{1}{3}\times\frac{5}5{}=\frac{5}{15}\)

\(\displaystyle \frac{2}{5}\times\frac{3}{3}=\frac{6}{15}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{5}{15}+\frac{6}{15}=\frac{11}{15}\)

\(\displaystyle \frac{1}{2}+\frac{2}{7}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{1}{2}\times\frac{7}{7}=\frac{7}{14}\)

\(\displaystyle \frac{2}{7}\times\frac{2}{2}=\frac{4}{14}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{7}{14}+\frac{4}{14}=\frac{11}{14}\)

\(\displaystyle \frac{3}{4}+\frac{1}{3}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{3}{4}\times\frac{3}{3}=\frac{9}{12}\)

\(\displaystyle \frac{1}{3}\times\frac{4}{4}=\frac{4}{12}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{9}{12}+\frac{4}{12}=\frac{13}{12}\)

\(\displaystyle \frac{13}{12}=1\frac{1}{12}\) because \(\displaystyle 12\) can go into \(\displaystyle 13\) one time, with one left over. 

\(\displaystyle \frac{5}{6}+\frac{2}{4}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{5}{6}\times\frac{2}{2}=\frac{10}{12}\)

\(\displaystyle \frac{2}{4}\times\frac{3}{3}=\frac{6}{12}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{10}{12}+\frac{6}{12}=\frac{16}{12}\)

\(\displaystyle \frac{16}{12}=1\frac{4}{12}=1\frac{1}{3}\) because \(\displaystyle 12\) and go into \(\displaystyle 16\) one time with four left over. \(\displaystyle \frac{4}{12}\) can be reduced to \(\displaystyle \frac{1}{3}\) by dividing the numerator and the denominator by \(\displaystyle 4\)

\(\displaystyle \frac{2}{3}+\frac{1}{7}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{2}{3}\times\frac{7}{7}=\frac{14}{21}\)

\(\displaystyle \frac{1}{7}\times\frac{3}{3}=\frac{3}{21}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{14}{21}+\frac{3}{21}=\frac{17}{21}\)

\(\displaystyle \frac{2}{3}+\frac{1}2{}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{2}{3}\times\frac{2}{2}=\frac{4}{6}\)

\(\displaystyle \frac{1}{2}\times\frac{}3{3}=\frac{3}{6}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{4}{6}+\frac{3}{6}=\frac{7}{6}\)

\(\displaystyle \frac{7}{6}=1\frac{1}{6}\) because \(\displaystyle 6\) can go into \(\displaystyle 7\) one time with \(\displaystyle \frac{1}{6}\) left over. 

\(\displaystyle \frac{1}{4}+\frac{4}{8}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{1}{4}\times\frac{2}{2}=\frac{2}{8}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{2}{8}+\frac{4}{8}=\frac{6}{8}\)

\(\displaystyle \frac{6}{8}\div\frac{2}{2}=\frac{3}{4}\)

\(\displaystyle \frac{4}6{+\frac{3}{8}}\)

In order to solve this problem, we first have to find common denominators. 

\(\displaystyle \frac{4}{6}\times\frac{4}{4}=\frac{16}{24}\)

\(\displaystyle \frac{3}{8}\times\frac{3}{3}=\frac{9}{24}\)

Now that we have common denominators, we can add the fractions. Remember, when we add and subtract fractions, we only add or subtract the numerator. 

\(\displaystyle \frac{16}{24}+\frac{9}{24}=\frac{25}{24}\)

\(\displaystyle \frac{25}{24}=1\frac{1}{24}\) because \(\displaystyle 24\) can go into \(\displaystyle 25\) one time with \(\displaystyle 1\) left over.