3(u+2)+u=+4(u-1)+10

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



3(u+2)+u=+4(u-1)+10

We move all terms to the left:

3(u+2)+u-(+4(u-1)+10)=0

We add all the numbers together, and all the variables

u+3(u+2)-(+4(u-1)+10)=0

We multiply parentheses

u+3u-(+4(u-1)+10)+6=0



We calculate terms in parentheses: -(+4(u-1)+10), so:

+4(u-1)+10

determiningTheFunctionDomain
4(u-1)+10

We multiply parentheses

4u-4+10

We add all the numbers together, and all the variables

4u+6

Back to the equation:

-(4u+6)



We add all the numbers together, and all the variables

4u-(4u+6)+6=0

We get rid of parentheses

4u-4u-6+6=0

We add all the numbers together, and all the variables

=0

u=0/1

u=0

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