2.   Solve :   

             (a)  \frac{1}{18} +\frac{1}{18}

             (b)   \frac{8}{15} +\frac{3}{15}

             (c)   \frac{7}{7} -\frac{5}{7}

             (d)  \frac{1}{22} +\frac{21}{22}

             (e)  \frac{12}{15} -\frac{7}{15}

             (f)   \frac{5}{8} +\frac{3}{8}

             (g)  1-\frac{2}{3}\left ( 1=\frac{3}{3} \right )

            ( h)  \frac{1}{4}+\frac{0}{4}

             (i)  3-\frac{12}{5}

Answers (1)

(a)   \frac{1}{18} +\frac{1}{18}\ =\ \frac{1+1}{18}\ =\ \frac{2}{18}\ =\ \frac{1}{9}

 (b)   \frac{8}{15} +\frac{3}{15}\ =\ \frac{8+3}{15}\ =\ \frac{11}{15}

  (c)   \frac{7}{7} -\frac{5}{7}\ =\ \frac{7-5}{7}\ =\ \frac{2}{7}

 (d)  \frac{1}{22} +\frac{21}{22}\ =\ \frac{1+21}{22}\ =\ \frac{22}{22}\ =\ 1

(e)  \frac{12}{15} -\frac{7}{15}\ =\ \frac{12-7}{15}\ =\ \frac{5}{15}\ =\ \frac{1}{ 3}

  (f)   \frac{5}{8} +\frac{3}{8}\ =\ \frac{5+3}{8}\ =\ \frac{8}{8}\ =\ 1

  (g)  1-\frac{2}{3}\ =\ \frac{3}{3}\ -\ \frac{2}{3}\ =\ \frac{3-2}{3}\ =\ \frac{1}{3}

   ( h)  \frac{1}{4}+\frac{0}{4}\ =\ \frac{1+0}{4}\ =\ \frac{1}{4}

    (i)  3-\frac{12}{5}\ =\ \frac{15}{5}\ -\ \frac{12}{5}\ =\ \frac{15-12}{5}\ =\ \frac{3}{5}

  

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