3(x-5)+(6x+1)=41

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Solution for 3(x-5)+(6x+1)=41 equation: 



3(x-5)+(6x+1)=41

We move all terms to the left:

3(x-5)+(6x+1)-(41)=0

We multiply parentheses

3x+(6x+1)-15-41=0

We get rid of parentheses

3x+6x+1-15-41=0

We add all the numbers together, and all the variables

9x-55=0

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

9x=55

x=55/9

x=6+1/9

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