question_answer 1)

                Work out the following divisions:                 (i) \[(10x-25)\div 5\]                 (ii) \[(10x-25)\div \,(2x-5)\]                 (iii) \[10y(6y+21)\,\div 5(2y+7)\]                 (iv) \[9{{x}^{2}}{{y}^{2}}\,(3z-24)\div \,27xy\,(z-8)\]                 (v) \[96abc(3a-12)\,(5b-30)\div \,144(a-4)\,(b-b)\].

Answer:

                (i) \[(10x-25)\div 5\]                 \[(10x-25)\div 5\] \[=\frac{10x-25}{5}\] \[=\frac{5(2x-5)}{5}\] \[=2x-5\]                 (ii) \[(10x-25)\div \,(2x-5)\]                         \[(10x-25)\div \,(2x-5)\] \[=\frac{10x-25}{2x-5}\] \[=\frac{5(2x-5)}{2x-5}\,=5\]                 (iii) \[10y(6y+21)\,\div 5(2y+7)\]                 \[10y(6y+21)\,\div 5(2y+7)\] \[=\frac{10y(6y+21)}{5(2y+7)}\] \[=\frac{10y\times 3(2y+7)}{5(2y+7)}\]                 (iv) \[9{{x}^{2}}{{y}^{2}}\,(3z-24)\div \,27xy\,(z-8)\]                 \[9{{x}^{2}}{{y}^{2}}\,(3z-24)\div \,27xy\,(z-8)\] \[=\frac{9{{x}^{2}}{{y}^{2}}(3z-24)}{27\,xy\,(z-8)}\] \[=\frac{9{{x}^{2}}{{y}^{2}}\times 3(z-8)}{27xy\,(z-8)}\]                 (v) \[96abc(3a-12)\,(5b-30)\div \,144(a-4)\,(b-b)\]. \[96abc(3a-12)\,(5b-30)\div \,144(a-4)\,(b-b)\] \[=\frac{96abc(3a-12)(5b-30)}{144(a-4)\,(b-6)}\] \[=\frac{96abc\times 3\,(a-4)\,\times 5(b-6)}{144(a-4)\,(b-6)}\] \[=10abc\].