6(3-x)+(-20)=10+3(9x+2)

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Solution for 6(3-x)+(-20)=10+3(9x+2) equation: 



6(3-x)+(-20)=10+3(9x+2)

We move all terms to the left:

6(3-x)+(-20)-(10+3(9x+2))=0

We add all the numbers together, and all the variables

6(-1x+3)-(10+3(9x+2))+(-20)=0

We add all the numbers together, and all the variables

6(-1x+3)-(10+3(9x+2))-20=0

We multiply parentheses

-6x-(10+3(9x+2))+18-20=0



We calculate terms in parentheses: -(10+3(9x+2)), so:

10+3(9x+2)

determiningTheFunctionDomain
3(9x+2)+10

We multiply parentheses

27x+6+10

We add all the numbers together, and all the variables

27x+16

Back to the equation:

-(27x+16)



We add all the numbers together, and all the variables

-6x-(27x+16)-2=0

We get rid of parentheses

-6x-27x-16-2=0

We add all the numbers together, and all the variables

-33x-18=0

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

-33x=18

x=18/-33

x=-6/11

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