[tex]\begin{gathered} a)\frac{3}{4} \\ b)\frac{5}{4} \\ c)\frac{7}{8} \\ d)\frac{5}{8} \end{gathered}[/tex]
1) Let's use the fraction circles then to solve this sum of fractions:
1) Note that since these fractions do not have the same denominator we can write out the following. Placing the LCM as the denominator\div
[tex]\begin{gathered} a)\frac{1}{2}+\frac{1}{4}=\frac{4\div2\times1+4\div4\times1}{4}=\frac{3}{4} \\ b)LCM(2,4)=4 \\ \frac{1}{2}+\frac{3}{4}=\frac{4\div2\times1}{4}+\frac{4\div4\times3}{4}=\frac{5}{4} \\ c)\text{LCM(4,}8)=8 \\ \frac{3}{4}+\frac{1}{8}=\frac{6+1}{8}=\frac{7}{8} \\ d)\text{LCM(}4,8)=8 \\ \frac{1}{4}+\frac{3}{8}=\frac{2+3}{8}=\frac{5}{8} \end{gathered}[/tex]
Note that for 'c' and "d" we have made it faster. Since this is the same principle.
