751+21(n-12)=649+31(n-10)

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Solution for 751+21(n-12)=649+31(n-10) equation: 



751+21(n-12)=649+31(n-10)

We move all terms to the left:

751+21(n-12)-(649+31(n-10))=0

We multiply parentheses

21n-(649+31(n-10))-252+751=0



We calculate terms in parentheses: -(649+31(n-10)), so:

649+31(n-10)

determiningTheFunctionDomain
31(n-10)+649

We multiply parentheses

31n-310+649

We add all the numbers together, and all the variables

31n+339

Back to the equation:

-(31n+339)



We add all the numbers together, and all the variables

21n-(31n+339)+499=0

We get rid of parentheses

21n-31n-339+499=0

We add all the numbers together, and all the variables

-10n+160=0

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

-10n=-160

n=-160/-10

n=+16

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