6(k+2)+1=4(k-1)

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Solution for 6(k+2)+1=4(k-1) equation: 



6(k+2)+1=4(k-1)

We move all terms to the left:

6(k+2)+1-(4(k-1))=0

We multiply parentheses

6k-(4(k-1))+12+1=0



We calculate terms in parentheses: -(4(k-1)), so:

4(k-1)

We multiply parentheses

4k-4

Back to the equation:

-(4k-4)



We add all the numbers together, and all the variables

6k-(4k-4)+13=0

We get rid of parentheses

6k-4k+4+13=0

We add all the numbers together, and all the variables

2k+17=0

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

2k=-17

k=-17/2

k=-8+1/2

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