Question

Qa. Divide [6 (90"b"- 300?be + 35ac1)) by (3 (3a4b-5ac)]
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Answer 1

Given:

Given numbers are:

$6(90b-300be+ac^1)$ and $3(3a^4b-5ac)$

To find:

We need to divide the given numbers.

Solution:

Given numbers are, $6(90b-300be+ac^1)$ and $3(3a^4b-5ac)$.

Now, dividing the given numbers,

$\Rightarrow \frac{6(90b-300be+ac^1)}{3(3a^4b-5ac)}$

Dividing the number 6 (present in numerator) with the number 3 (present in denominator),

$\Rightarrow \frac{2(90b-300be+ac^1)}{1(3a^4b-5ac)}$

Simplifying the terms which are present in the numerator,

$\Rightarrow \frac{180b-600be+2ac^1}{1(3a^4b-5ac)}$

Simplifying the terms which are present in the denominator,

$\Rightarrow \frac{180b-600be+2ac^1}{3a^4b-5ac}$

We know that, $a^1=1$.

$\Rightarrow \frac{180b-600be+2ac}{3a^4b-5ac}$

From the above expression, it is clear that, there are no common terms in numerator and the denominator.

So, the above terms cannot be simplified further.

Therefore, the value of  $\frac{6(90b-300be+ac^1)}{3(3a^4b-5ac)}= \frac{180b-600be+2ac}{3a^4b-5ac}$.