Question

divide
$(6y {2} + 2y) \div 2y$
​

Answer 1

$\sf\large \bold{Answer:}$

$\sf \: 3x + 1$

$\sf\large \bold{Step \: by\:step\: explanation:}$

$\sf \bold{Divide \: \: 6 {y}^{2} + 2y \: \: by \: 2y }$

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⟹⠀$\sf \dfrac{ 6{y}^{2} + 2y }{2y}$

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Take 2y as common from the numerator.

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⟹⠀$\sf \dfrac{2y(3y + 1)}{2y}$

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2y divided by 2y gives 1

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⟹⠀$\sf \dfrac{{\not{2\not y}}(3x+1)}{\not{2\not y}}$

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⟹⠀$\sf \: 3x + 1$

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[ How did we divide :

First factorise the dividend and divisor if possible. Or Numerator and denominator, Then cancel the highest common factor.

We will get the quotient

let us take an simple example

ex: 12 ÷ 2

= 2 × 2 × 3 / 2

= 2 × 3

= 6

we can do same thing with expressions ]

Answer 2

$\huge\mathcal\purple{AnSwEr!}$

$\small{ }$

$\bold{To~find:-}$

$\large\mathtt{(6y^{2}+2y)÷2y}$

i.e., $\large\mathtt{{\frac{(6y^{2}+2y)}{2y}}}$

Taking 2y as a common factor, we get:

$\large\mathtt{{\frac{{\cancel{2y}}(3y+1)}{{\cancel{2y}}}}}$

$\large\implies\mathtt{{\underline{\underline{3y+1}}}}$

$\small{ }$

$\large\therefore$ Required answer:

$\large\mathtt{{\underline{\underline{3y+1}}}}$

$\small{ }$

$\huge\mathrm\purple{Høpe~it~helps..!!}$