4(3u-1)+5u+1=6(u+2)-4u

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Solution for 4(3u-1)+5u+1=6(u+2)-4u equation: 



4(3u-1)+5u+1=6(u+2)-4u

We move all terms to the left:

4(3u-1)+5u+1-(6(u+2)-4u)=0

We add all the numbers together, and all the variables

5u+4(3u-1)-(6(u+2)-4u)+1=0

We multiply parentheses

5u+12u-(6(u+2)-4u)-4+1=0



We calculate terms in parentheses: -(6(u+2)-4u), so:

6(u+2)-4u

We add all the numbers together, and all the variables

-4u+6(u+2)

We multiply parentheses

-4u+6u+12

We add all the numbers together, and all the variables

2u+12

Back to the equation:

-(2u+12)



We add all the numbers together, and all the variables

17u-(2u+12)-3=0

We get rid of parentheses

17u-2u-12-3=0

We add all the numbers together, and all the variables

15u-15=0

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

15u=15

u=15/15

u=1

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