5(x+3)+4(x-5)=1+6(x-2)

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Solution for 5(x+3)+4(x-5)=1+6(x-2) equation: 



5(x+3)+4(x-5)=1+6(x-2)

We move all terms to the left:

5(x+3)+4(x-5)-(1+6(x-2))=0

We multiply parentheses

5x+4x-(1+6(x-2))+15-20=0



We calculate terms in parentheses: -(1+6(x-2)), so:

1+6(x-2)

determiningTheFunctionDomain
6(x-2)+1

We multiply parentheses

6x-12+1

We add all the numbers together, and all the variables

6x-11

Back to the equation:

-(6x-11)



We add all the numbers together, and all the variables

9x-(6x-11)-5=0

We get rid of parentheses

9x-6x+11-5=0

We add all the numbers together, and all the variables

3x+6=0

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

3x=-6

x=-6/3

x=-2

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