k+2(k-3)=22(4k+3)+k

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Solution for k+2(k-3)=22(4k+3)+k equation: 



k+2(k-3)=22(4k+3)+k

We move all terms to the left:

k+2(k-3)-(22(4k+3)+k)=0

We multiply parentheses

k+2k-(22(4k+3)+k)-6=0



We calculate terms in parentheses: -(22(4k+3)+k), so:

22(4k+3)+k

We add all the numbers together, and all the variables

k+22(4k+3)

We multiply parentheses

k+88k+66

We add all the numbers together, and all the variables

89k+66

Back to the equation:

-(89k+66)



We add all the numbers together, and all the variables

3k-(89k+66)-6=0

We get rid of parentheses

3k-89k-66-6=0

We add all the numbers together, and all the variables

-86k-72=0

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

-86k=72

k=72/-86

k=-36/43

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