Question

(6) \\(\frac{9}{10} + \frac{3}{20} = \frac{18}{20} + \frac{3}{20} = 1\frac{1}{20}\\)
(7) \\(\frac{1}{12} + \frac{1}{28} = \\)
(8) \\(\frac{5}{24} + \frac{1}{30} = \\)
(9) \\(\frac{5}{17} + \frac{1}{34} = \\)
(10) \\(\frac{1}{15} + \frac{2}{35} = \\)
(11) \\(\frac{1}{20} + \frac{3}{40} = \\)

Explanation:

Step1: Solve \(\frac{1}{12}+\frac{1}{28}\)

Find the least common denominator (LCD) of 12 and 28. Prime factors of 12: \(2^2\times3\), prime factors of 28: \(2^2\times7\). So LCD is \(2^2\times3\times7 = 84\).
Rewrite fractions: \(\frac{1\times7}{12\times7}=\frac{7}{84}\), \(\frac{1\times3}{28\times3}=\frac{3}{84}\).
Add: \(\frac{7 + 3}{84}=\frac{10}{84}=\frac{5}{42}\).

Step2: Solve \(\frac{5}{24}+\frac{1}{30}\)

Prime factors of 24: \(2^3\times3\), prime factors of 30: \(2\times3\times5\). LCD is \(2^3\times3\times5 = 120\).
Rewrite: \(\frac{5\times5}{24\times5}=\frac{25}{120}\), \(\frac{1\times4}{30\times4}=\frac{4}{120}\).
Add: \(\frac{25 + 4}{120}=\frac{29}{120}\).

Step3: Solve \(\frac{5}{17}+\frac{1}{34}\)

LCD of 17 and 34 is 34.
Rewrite: \(\frac{5\times2}{17\times2}=\frac{10}{34}\), \(\frac{1}{34}\) remains.
Add: \(\frac{10 + 1}{34}=\frac{11}{34}\).

Step4: Solve \(\frac{1}{15}+\frac{2}{35}\)

Prime factors of 15: \(3\times5\), prime factors of 35: \(5\times7\). LCD is \(3\times5\times7 = 105\).
Rewrite: \(\frac{1\times7}{15\times7}=\frac{7}{105}\), \(\frac{2\times3}{35\times3}=\frac{6}{105}\).
Add: \(\frac{7 + 6}{105}=\frac{13}{105}\).

Step5: Solve \(\frac{1}{20}+\frac{3}{40}\)

LCD of 20 and 40 is 40.
Rewrite: \(\frac{1\times2}{20\times2}=\frac{2}{40}\), \(\frac{3}{40}\) remains.
Add: \(\frac{2 + 3}{40}=\frac{5}{40}=\frac{1}{8}\).

Answer:

(7) \(\boldsymbol{\frac{5}{42}}\)
(8) \(\boldsymbol{\frac{29}{120}}\)
(9) \(\boldsymbol{\frac{11}{34}}\)
(10) \(\boldsymbol{\frac{13}{105}}\)
(11) \(\boldsymbol{\frac{1}{8}}\)