If needed, here is an example of using factoring to add the fractions in a):
\(\dfrac{3}{2\cdot{2}\cdot{2}}+\dfrac{1}{2\cdot{3}}=\dfrac{3\cdot{3}}{2\cdot{2}\cdot{2}\cdot{3}}+\dfrac{1\cdot{2}\cdot{2}}{2\cdot{3}\cdot{2}\cdot{2}}=\dfrac{9+4}{2\cdot{2}\cdot{2}\cdot{3}}=\dfrac{13}{24}\)
It may be helpful to explain in writing how to add the fractions for each of the problems. Below are a set of procedures for problem a) to get you started:
- Factor each of the two denominators: 6 and 8.
- Rewrite the denominators with the prime factorizations.
- Find the LCM of the denominators by looking at which number has the greatest number of each factor. In this case, three 2’s and one 3.
- Multiply the numerator and denominator of each fraction by the missing factors of the LCM. The first fraction is multiplied by \(\dfrac{3}{3}\). The second fraction is multiplied by \(\dfrac{2\cdot{2}}{2\cdot{2}}\).
- Multiply the factors of each numerator and take their sum. Multiply the factor of the denominator (LCM). Result is the sum of the two fractions.
Solutions:
a) \(\dfrac{13}{24}\) b) \(\dfrac{7}{8}\) c) \(\dfrac{127}{336}\) d) \(\dfrac{8}{15}\)