6(k-3)+2(k+5)=7(k-1)

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Solution for 6(k-3)+2(k+5)=7(k-1) equation: 



6(k-3)+2(k+5)=7(k-1)

We move all terms to the left:

6(k-3)+2(k+5)-(7(k-1))=0

We multiply parentheses

6k+2k-(7(k-1))-18+10=0



We calculate terms in parentheses: -(7(k-1)), so:

7(k-1)

We multiply parentheses

7k-7

Back to the equation:

-(7k-7)



We add all the numbers together, and all the variables

8k-(7k-7)-8=0

We get rid of parentheses

8k-7k+7-8=0

We add all the numbers together, and all the variables

k-1=0

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

k=1

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