3(k+4)=5(k-6)+32

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Solution for 3(k+4)=5(k-6)+32 equation: 



3(k+4)=5(k-6)+32

We move all terms to the left:

3(k+4)-(5(k-6)+32)=0

We multiply parentheses

3k-(5(k-6)+32)+12=0



We calculate terms in parentheses: -(5(k-6)+32), so:

5(k-6)+32

We multiply parentheses

5k-30+32

We add all the numbers together, and all the variables

5k+2

Back to the equation:

-(5k+2)



We get rid of parentheses

3k-5k-2+12=0

We add all the numbers together, and all the variables

-2k+10=0

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

-2k=-10

k=-10/-2

k=+5

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