4(4x+1)=3x-4(6-x)+8=

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Solution for 4(4x+1)=3x-4(6-x)+8= equation: 



4(4x+1)=3x-4(6-x)+8=

We move all terms to the left:

4(4x+1)-(3x-4(6-x)+8)=0

We add all the numbers together, and all the variables

4(4x+1)-(3x-4(-1x+6)+8)=0

We multiply parentheses

16x-(3x-4(-1x+6)+8)+4=0



We calculate terms in parentheses: -(3x-4(-1x+6)+8), so:

3x-4(-1x+6)+8

We multiply parentheses

3x+4x-24+8

We add all the numbers together, and all the variables

7x-16

Back to the equation:

-(7x-16)



We get rid of parentheses

16x-7x+16+4=0

We add all the numbers together, and all the variables

9x+20=0

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

9x=-20

x=-20/9

x=-2+2/9

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