(3x=1)+(2x-1)+(5x-4)=46

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Solution for (3x=1)+(2x-1)+(5x-4)=46 equation: 



(3x=1)+(2x-1)+(5x-4)=46

We move all terms to the left:

(3x-(1)+(2x-1)+(5x-4))=0



We calculate terms in parentheses: +(3x-1+(2x-1)+(5x-4)), so:

3x-1+(2x-1)+(5x-4)

determiningTheFunctionDomain
3x+(2x-1)+(5x-4)-1

We get rid of parentheses

3x+2x+5x-1-4-1

We add all the numbers together, and all the variables

10x-6

Back to the equation:

+(10x-6)



We get rid of parentheses

10x-6=0

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

10x=6

x=6/10

x=3/5

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