10(2x-4-4x)+6x=8(4-x)-8x

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Solution for 10(2x-4-4x)+6x=8(4-x)-8x equation: 



10(2x-4-4x)+6x=8(4-x)-8x

We move all terms to the left:

10(2x-4-4x)+6x-(8(4-x)-8x)=0

We add all the numbers together, and all the variables

10(-2x-4)+6x-(8(-1x+4)-8x)=0

We add all the numbers together, and all the variables

6x+10(-2x-4)-(8(-1x+4)-8x)=0

We multiply parentheses

6x-20x-(8(-1x+4)-8x)-40=0



We calculate terms in parentheses: -(8(-1x+4)-8x), so:

8(-1x+4)-8x

We add all the numbers together, and all the variables

-8x+8(-1x+4)

We multiply parentheses

-8x-8x+32

We add all the numbers together, and all the variables

-16x+32

Back to the equation:

-(-16x+32)



We add all the numbers together, and all the variables

-14x-(-16x+32)-40=0

We get rid of parentheses

-14x+16x-32-40=0

We add all the numbers together, and all the variables

2x-72=0

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

2x=72

x=72/2

x=36

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