Evaluate:
(2×14)−(−187×−715)\Big(2 \times \dfrac{1}{4}\Big) - \Big(\dfrac{-18}{7} \times \dfrac{-7}{15}\Big)(2×41)−(7−18×15−7)
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(2×11×4)−(−18×−77×15)=(24)−(126105)=(12)−(65)\Big(\dfrac{2 \times 1}{1 \times 4}\Big) - \Big(\dfrac{-18 \times -7}{7 \times 15}\Big) \\[1em] =\Big(\dfrac{2}{4}\Big) - \Big(\dfrac{126}{105}\Big) \\[1em] =\Big(\dfrac{1}{2}\Big) - \Big(\dfrac{6}{5}\Big) \\[1em](1×42×1)−(7×15−18×−7)=(42)−(105126)=(21)−(56) LCM of 2 and 5 is 2 x 5 = 10 =(1×52×5)−(6×25×2)=(510)−(1210)=(5−1210)=(−710)=\Big(\dfrac{1 \times 5}{2 \times 5}\Big) - \Big(\dfrac{6 \times 2}{5 \times 2}\Big)\\[1em] =\Big(\dfrac{5}{10}\Big) - \Big(\dfrac{12}{10}\Big)\\[1em] =\Big(\dfrac{5 - 12}{10}\Big)\\[1em] =\Big(\dfrac{-7}{10}\Big)=(2×51×5)−(5×26×2)=(105)−(1012)=(105−12)=(10−7)
Hence, (2×14)−(−187×−715)=(−710)\Big(2 \times \dfrac{1}{4}\Big) - \Big(\dfrac{-18}{7} \times \dfrac{-7}{15}\Big) = \Big(\dfrac{-7}{10}\Big)(2×41)−(7−18×15−7)=(10−7)
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Multiply:
−710\dfrac{-7}{10}10−7 and −815\dfrac{-8}{15}15−8
(2−3×54)+(59×3−10)\Big(\dfrac{2}{-3} \times \dfrac{5}{4}\Big) + \Big(\dfrac{5}{9} \times \dfrac{3}{-10}\Big)(−32×45)+(95×−103)
(−5×215)−(−6×29)\Big(-5 \times \dfrac{2}{15}\Big) - \Big(-6 \times \dfrac{2}{9}\Big)(−5×152)−(−6×92)
(85×−32)+(−310×916)\Big(\dfrac{8}{5} \times \dfrac{-3}{2}\Big) + \Big(\dfrac{-3}{10} \times \dfrac{9}{16}\Big)(58×2−3)+(10−3×169)
