6(k+4)-(-2)(k+5)+8=3(k+1)-9

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Solution for 6(k+4)-(-2)(k+5)+8=3(k+1)-9 equation: 



6(k+4)-(-2)(k+5)+8=3(k+1)-9

We move all terms to the left:

6(k+4)-(-2)(k+5)+8-(3(k+1)-9)=0

We multiply parentheses

6k-(-2)(k+5)-(3(k+1)-9)+24+8=0

We multiply parentheses ..

6k-(-2k-10)-(3(k+1)-9)+24+8=0



We calculate terms in parentheses: -(3(k+1)-9), so:

3(k+1)-9

We multiply parentheses

3k+3-9

We add all the numbers together, and all the variables

3k-6

Back to the equation:

-(3k-6)



We add all the numbers together, and all the variables

6k-(-2k-10)-(3k-6)+32=0

We get rid of parentheses

6k+2k-3k+10+6+32=0

We add all the numbers together, and all the variables

5k+48=0

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

5k=-48

k=-48/5

k=-9+3/5

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