540=113+(19x)+(100-x)+(19x)+124+(98+3x)

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Solution for 540=113+(19x)+(100-x)+(19x)+124+(98+3x) equation: 



540=113+(19x)+(100-x)+(19x)+124+(98+3x)

We move all terms to the left:

540-(113+(19x)+(100-x)+(19x)+124+(98+3x))=0

We add all the numbers together, and all the variables

-(113+19x+(-1x+100)+19x+124+(3x+98))+540=0



We calculate terms in parentheses: -(113+19x+(-1x+100)+19x+124+(3x+98)), so:

113+19x+(-1x+100)+19x+124+(3x+98)

determiningTheFunctionDomain
19x+(-1x+100)+19x+(3x+98)+113+124

We add all the numbers together, and all the variables

38x+(-1x+100)+(3x+98)+237

We get rid of parentheses

38x-1x+3x+100+98+237

We add all the numbers together, and all the variables

40x+435

Back to the equation:

-(40x+435)



We get rid of parentheses

-40x-435+540=0

We add all the numbers together, and all the variables

-40x+105=0

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

-40x=-105

x=-105/-40

x=2+5/8

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