21(7j-3)+28(10j-20)=84-12(2j-5)

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Solution for 21(7j-3)+28(10j-20)=84-12(2j-5) equation: 



21(7j-3)+28(10j-20)=84-12(2j-5)

We move all terms to the left:

21(7j-3)+28(10j-20)-(84-12(2j-5))=0

We multiply parentheses

147j+280j-(84-12(2j-5))-63-560=0



We calculate terms in parentheses: -(84-12(2j-5)), so:

84-12(2j-5)

determiningTheFunctionDomain
-12(2j-5)+84

We multiply parentheses

-24j+60+84

We add all the numbers together, and all the variables

-24j+144

Back to the equation:

-(-24j+144)



We add all the numbers together, and all the variables

427j-(-24j+144)-623=0

We get rid of parentheses

427j+24j-144-623=0

We add all the numbers together, and all the variables

451j-767=0

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

451j=767

j=767/451

j=1+316/451

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