Question

answer the solution quickly...​

Answer 1

the answer of the equation is 25

Answer 2

Given :

$\\ :\normalsize\boxed{\bf \dfrac{x+3}{7}-\dfrac{2x-5}{3}=\dfrac{3x-5}{5}-25}$

To FinD :

The value of x.

Solution :

Analysis :

We need to find the value of x by using the LCM of the numbers and by using the evaluation signs.

Explanation :

$\\ :\implies\normalsize{\sf \dfrac{x+3}{7}-\dfrac{2x-5}{3}=\dfrac{3x-5}{5}-25}$

Taking LCM of 7,3 = 21 in LHS and 5,1 = 5 in RHS,

$\\ :\implies\normalsize{\sf \dfrac{3(x+3)-7(2x-5)}{21}=\dfrac{3x-5-5(25)}{5}}$

Dividing the LCM by denominator and multiplying the numerator by the quotient of the division of the denominator,

$\\ :\implies\normalsize{\sf \dfrac{3x+9-14x+35}{21}=\dfrac{3x-5-125}{5}}$

Organising the numbers as per their signs,

$\\ :\implies\normalsize{\sf \dfrac{3x-14x+9+35}{21}=\dfrac{3x-5-125}{5}}$

Evaluating the numbers as per their signs,

$\\ :\implies\normalsize{\sf \dfrac{-11x+44}{21}=\dfrac{3x-130}{5}}$

By cross multiplying,

$\\ :\implies\normalsize{\sf 5(-11x+44)=21(3x-130)}$

After multiplying the numbers,

$\\ :\implies\normalsize{\sf -55x+220=63x-2730}$

Transposing the numbers with x to RHS and normal numbers to LHS,

$\\ :\implies\normalsize{\sf 220+2730=63x+55x}$

After evaluation,

$\\ :\implies\normalsize{\sf 2950=118x}$

$\\ :\implies\normalsize{\sf \dfrac{2950}{118}=x}$

$\\ :\implies\normalsize{\sf \cancel{\dfrac{2950}{118}}=x}$

$\\ :\implies\normalsize{\sf 25=x}$

$\\ \large\therefore\boxed{\bf x=25.}$