2(u+3)=3(1-3u)+23

Solution for 2(u+3)=3(1-3u)+23 equation:

2(u+3)=3(1-3u)+23

We move all terms to the left:

2(u+3)-(3(1-3u)+23)=0

We add all the numbers together, and all the variables

2(u+3)-(3(-3u+1)+23)=0

We multiply parentheses

2u-(3(-3u+1)+23)+6=0


We calculate terms in parentheses: -(3(-3u+1)+23), so:

3(-3u+1)+23

We multiply parentheses

-9u+3+23

We add all the numbers together, and all the variables

-9u+26

Back to the equation:

-(-9u+26)


We get rid of parentheses

2u+9u-26+6=0

We add all the numbers together, and all the variables

11u-20=0

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

11u=20

u=20/11

u=1+9/11