2(k-5)+3(k-2)=8+7(k-4)

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Solution for 2(k-5)+3(k-2)=8+7(k-4) equation: 



2(k-5)+3(k-2)=8+7(k-4)

We move all terms to the left:

2(k-5)+3(k-2)-(8+7(k-4))=0

We multiply parentheses

2k+3k-(8+7(k-4))-10-6=0



We calculate terms in parentheses: -(8+7(k-4)), so:

8+7(k-4)

determiningTheFunctionDomain
7(k-4)+8

We multiply parentheses

7k-28+8

We add all the numbers together, and all the variables

7k-20

Back to the equation:

-(7k-20)



We add all the numbers together, and all the variables

5k-(7k-20)-16=0

We get rid of parentheses

5k-7k+20-16=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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