((3x+19)/4)+((x+53)/7)=15

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Solution for ((3x+19)/4)+((x+53)/7)=15 equation: 



((3x+19)/4)+((x+53)/7)=15

We move all terms to the left:

((3x+19)/4)+((x+53)/7)-(15)=0

We calculate fractions


(21x+133)/()+(4x+212)/()-15=0

We multiply all the terms by the denominator


(21x+133)+(4x+212)-15*()=0

We add all the numbers together, and all the variables

(21x+133)+(4x+212)=0

We get rid of parentheses

21x+4x+133+212=0

We add all the numbers together, and all the variables

25x+345=0

We move all terms containing x to the left, all other terms to the right

25x=-345

x=-345/25

x=-13+4/5

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