4-5(20x-4)=128-6(3-12x)

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Solution for 4-5(20x-4)=128-6(3-12x) equation: 



4-5(20x-4)=128-6(3-12x)

We move all terms to the left:

4-5(20x-4)-(128-6(3-12x))=0

We add all the numbers together, and all the variables

-5(20x-4)-(128-6(-12x+3))+4=0

We multiply parentheses

-100x-(128-6(-12x+3))+20+4=0



We calculate terms in parentheses: -(128-6(-12x+3)), so:

128-6(-12x+3)

determiningTheFunctionDomain
-6(-12x+3)+128

We multiply parentheses

72x-18+128

We add all the numbers together, and all the variables

72x+110

Back to the equation:

-(72x+110)



We add all the numbers together, and all the variables

-100x-(72x+110)+24=0

We get rid of parentheses

-100x-72x-110+24=0

We add all the numbers together, and all the variables

-172x-86=0

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

-172x=86

x=86/-172

x=-1/2

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