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\((\frac{8}{-15}+\frac{4}{-3})=\,?\)A. \(\frac{28}{15}\)B. \(\frac{-28}{15}\)C. \(\frac{-4}{5}\)D. \(\frac{-4}{15}\) |
Answer» \(\frac{8}{-15}=\frac{8\times-1}{-15\times-1}=\frac{-8}{15}\) And, \(\frac{4}{-3}=\frac{4\times-1}{-3\times-1}=\frac{-4}{3}\) \(\Rightarrow\) \(\frac{8}{-15}+\frac{4}{-3}=\frac{-8}{15}+\frac{-4}{3}\) \(\Rightarrow\) \(\frac{-8\times3+(-4)\times15}{45}\) \(\Rightarrow\) \(\frac{-24-60}{45}\) \(=\frac{-84}{45}=\frac{-84\div3}{45\div3}=\frac{-28}{15}\) |
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