(x-3)-2(x-5)=6+(x-2)

Solution for (x-3)-2(x-5)=6+(x-2) equation:

(x-3)-2(x-5)=6+(x-2)

We move all terms to the left:

(x-3)-2(x-5)-(6+(x-2))=0

We multiply parentheses

(x-3)-2x-(6+(x-2))+10=0

We get rid of parentheses

x-2x-(6+(x-2))-3+10=0


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

6+(x-2)

determiningTheFunctionDomain
(x-2)+6

We get rid of parentheses

x-2+6

We add all the numbers together, and all the variables

x+4

Back to the equation:

-(x+4)


We add all the numbers together, and all the variables

-1x-(x+4)+7=0

We get rid of parentheses

-1x-x-4+7=0

We add all the numbers together, and all the variables

-2x+3=0

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

-2x=-3

x=-3/-2

x=1+1/2