7(3+2k)=2+3(4k+5)

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Solution for 7(3+2k)=2+3(4k+5) equation: 



7(3+2k)=2+3(4k+5)

We move all terms to the left:

7(3+2k)-(2+3(4k+5))=0

We add all the numbers together, and all the variables

7(2k+3)-(2+3(4k+5))=0

We multiply parentheses

14k-(2+3(4k+5))+21=0



We calculate terms in parentheses: -(2+3(4k+5)), so:

2+3(4k+5)

determiningTheFunctionDomain
3(4k+5)+2

We multiply parentheses

12k+15+2

We add all the numbers together, and all the variables

12k+17

Back to the equation:

-(12k+17)



We get rid of parentheses

14k-12k-17+21=0

We add all the numbers together, and all the variables

2k+4=0

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

2k=-4

k=-4/2

k=-2

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