Divide the following:$96abc(3a-12)(5b-30) \div 144(a-4)(b-6)$

Given :

The given expression is $96abc(3a-12)(5b-30) \div 144(a-4)(b-6)$.

To do :

We have to divide the given expressions.

Solution :

$3a-12$ can be written as,

$3a-12=3(a-4)$   (Taking 3 common)

$5b-30$ can be written as,

$5b-30 = 5(b-6)$   (Taking 5 common)

Therefore,

$96abc(3a-12)(5b-30) ÷ 144(a-4)(b-6) = \frac{96abc(3a-12)(5b-30)}{144(a-4)(b-6)}$

                                                           $= \frac{96abc \times 3(a-4) \times 5(b-6)}{144(a-4)(b-6)}$

                                                          $ = \frac{(96 \times 3 \times 5 abc)}{144}$

                                                           $= \frac{8 \times 3 \times 5 abc}{12}$

                                                           $= \frac{2 \times 3 \times 5 abc}{3}$

                                                           $= 10abc$

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Updated on: 10-Oct-2022

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