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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-12x-(10(x-3)-10)+18+20=0



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

10(x-3)-10

We multiply parentheses

10x-30-10

We add all the numbers together, and all the variables

10x-40

Back to the equation:

-(10x-40)



We add all the numbers together, and all the variables

-12x-(10x-40)+38=0

We get rid of parentheses

-12x-10x+40+38=0

We add all the numbers together, and all the variables

-22x+78=0

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

-22x=-78

x=-78/-22

x=3+6/11

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