Fill in the missing fractions: |
(a) \[\frac{7}{10}-=\frac{3}{10}\] |
(b) \[-\frac{3}{21}=\frac{5}{21}\] |
(c) \[-\frac{3}{6}=\frac{3}{6}\] |
(d) \[+\frac{5}{27}=\frac{12}{27}\] |
Answer:
(a) Here, \[\frac{7}{10}-=\frac{3}{10}\] Subtracting \[\frac{7}{10}\] from both sides, we get \[\frac{7}{10}-=\frac{3}{10}=\frac{7}{10}-\frac{3}{10}\] Or \[\frac{7}{10}-\frac{3}{10}-=\frac{7-3}{10}=\frac{4}{10}\] \[\therefore \] \[\frac{7}{10}-\frac{4}{10}=\frac{3}{10}\] (b) \[-\frac{3}{21}=\frac{5}{21}\] Or \[=\frac{5}{21}+\frac{3}{21}\] Or \[=\frac{5+3}{21}\] Or \[=\frac{8}{21}\] \[\therefore \] \[\frac{8}{21}-\frac{3}{21}=\frac{5}{21}\] (c) \[-\frac{3}{6}=\frac{3}{6}\] \[=\frac{3}{6}+\frac{3}{6}\] \[=\frac{3+3}{6}\] \[=\frac{6}{6}(=1)\] \[\therefore \] \[1-\frac{3}{6}=\frac{3}{6}\] (d) \[+\frac{5}{27}=\frac{12}{27}\] or \[=\frac{12}{27}-\frac{5}{27}\] or \[=\frac{7}{27}\] \[\therefore \] \[\frac{7}{27}+\frac{5}{27}=\frac{12}{27}\].
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