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# Verify the following: (i) $\frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$ (ii) $\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$ (iii) $3+\frac{-7}{12}=\frac{-7}{12}+3$ (iv) $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

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Solution

## 1. LHS = $\frac{-12}{5}+\frac{2}{7}$ LCM of 5 and 7 is 35. $\frac{-12×7}{5×7}+\frac{2×5}{7×5}=\frac{-84}{35}+\frac{10}{35}=\frac{-84+10}{35}=\frac{-74}{35}$ RHS = $\frac{2}{7}+\frac{-12}{5}$ LCM of 5 and 7 is 35. $\frac{2×5}{7×5}+\frac{-12×7}{5×7}=\frac{10}{35}+\frac{-84}{35}=\frac{10-84}{35}=\frac{-74}{35}$ ∴ $\frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$ 2. â€‹LHS = $\frac{-5}{8}+\frac{-9}{13}$ LCM of 8 and 13 is 104. $\frac{-5×13}{8×13}+\frac{-9×8}{13×8}=\frac{-65}{104}+\frac{-72}{104}=\frac{-65+\left(-72\right)}{104}=\frac{-65-72}{104}=\frac{-137}{104}$ RHS = $\frac{-9}{13}+\frac{-5}{8}$ LCM of 13 and 8 is 104. $\frac{-9×8}{13×8}+\frac{-5×13}{8×13}=\frac{-72}{104}+\frac{-65}{104}=\frac{-72-65}{104}=\frac{-137}{104}$ ∴ $\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$ 3. â€‹LHS = $\frac{3}{1}+\frac{-7}{12}$ LCM of 1 and 12 is 12. $\frac{3×12}{1×12}+\frac{-7×1}{12×1}=\frac{36}{12}+\frac{-7}{12}=\frac{36+\left(-7\right)}{12}=\frac{36-7}{12}=\frac{29}{12}$ RHS = $\frac{-7}{12}+\frac{3}{1}$ LCM of 12 and 1 is 12. $\frac{-7×1}{12×1}+\frac{3×12}{1×12}=\frac{-7}{12}+\frac{36}{12}=\frac{-7+36}{12}=\frac{29}{12}$ ∴ $3+\frac{-7}{12}=\frac{-7}{12}+3$ 4. LHS = â€‹$\frac{2}{-7}+\frac{12}{-35}$ We will write the given numbers with positive denominators. $\frac{2}{-7}=\frac{2×\left(-1\right)}{-7×\left(-1\right)}=\frac{-2}{7}$ and $\frac{12}{-35}=\frac{12×\left(-1\right)}{-35×\left(-1\right)}=\frac{-12}{35}$ LCM of 7 and 35 is 35. $\frac{-2×5}{7×5}+\frac{-12×1}{35×1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+\left(-12\right)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$ RHS = $\frac{12}{-35}+\frac{2}{-7}$ We will write the given numbers with positive denominators. $\frac{12}{-35}=\frac{12×\left(-1\right)}{-35×\left(-1\right)}=\frac{-12}{35}$ and $\frac{2}{-7}=\frac{2×\left(-1\right)}{-7×\left(-1\right)}=\frac{-2}{7}$ LCM of 35 and 7 is 35. $\frac{-2×5}{7×5}+\frac{-12×1}{35×1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+\left(-12\right)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$ ∴â€‹ $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

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