Question

What will be the answer when we solving this equation?(Try to send proper explanation)
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Answer 1

Given :

$\normalsize\boxed{\bf\dfrac{x - 14}{5} + \dfrac{x - 0}{3} + \dfrac{x - 0}{6}=0}$

To FinD :

Evaluation.

Solution :

Analysis :

By using the symbols given and the fractions we have to evaluate it as per the sum.

Explanation :

$\\ :\implies\normalsize\sf\dfrac{x - 14}{5} + \dfrac{x - 0}{3} + \dfrac{x - 0}{6}=0$

By taking LCM of 5, 3 and 6 we get 30,

Then by dividing the denominator and multiplying the quotient of the denominator with the numerator we get,

$\\ :\implies\normalsize\sf\dfrac{6(x - 14)+10(x-0)+5(x-0)}{30}=0$

Then by multiplying the variants with the numbers beside them we get,

$\\ :\implies\normalsize\sf\dfrac{6x-84+10x-0+5x-0}{30}=0$

Now transposing the denominator 30 to RHS,

$\\ :\implies\normalsize\sf6x-84+10x-0+5x-0=0\times30$

$\\ :\implies\normalsize\sf6x-84+10x-0+5x-0=0$

Now evaluating as per the signs,

$\\ :\implies\normalsize\sf6x+10x+5x-84-0-0=0$

$\\ :\implies\normalsize\sf21x-84-0=0$

$\\ :\implies\normalsize\sf21x-84=0$

Now transposing -84 to RHS,

$\\ :\implies\normalsize\sf21x=0+84$

$\\ :\implies\normalsize\sf21x=84$

$\\ :\implies\normalsize\sf x=\dfrac{84}{21}$

$\\ :\implies\normalsize\sf x=\cancel{\dfrac{84}{21}}$

$\\ \normalsize\therefore\boxed{\bf{x=4.}}$

The value of x is 4.