12x-2(4-x)+4-(1-x)=3(x+2)-4(x-2)+10

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Solution for 12x-2(4-x)+4-(1-x)=3(x+2)-4(x-2)+10 equation: 



12x-2(4-x)+4-(1-x)=3(x+2)-4(x-2)+10

We move all terms to the left:

12x-2(4-x)+4-(1-x)-(3(x+2)-4(x-2)+10)=0

We add all the numbers together, and all the variables

12x-2(-1x+4)-(-1x+1)-(3(x+2)-4(x-2)+10)+4=0

We multiply parentheses

12x+2x-(-1x+1)-(3(x+2)-4(x-2)+10)-8+4=0

We get rid of parentheses

12x+2x+1x-(3(x+2)-4(x-2)+10)-1-8+4=0



We calculate terms in parentheses: -(3(x+2)-4(x-2)+10), so:

3(x+2)-4(x-2)+10

We multiply parentheses

3x-4x+6+8+10

We add all the numbers together, and all the variables

-1x+24

Back to the equation:

-(-1x+24)



We add all the numbers together, and all the variables

15x-(-1x+24)-5=0

We get rid of parentheses

15x+1x-24-5=0

We add all the numbers together, and all the variables

16x-29=0

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

16x=29

x=29/16

x=1+13/16

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