8(I+5)=4(2x+1)+36

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Solution for 8(I+5)=4(2x+1)+36 equation: 



8(+5)=4(2I+1)+36

We move all terms to the left:

8(+5)-(4(2I+1)+36)=0

We add all the numbers together, and all the variables

-(4(2I+1)+36)+85=0



We calculate terms in parentheses: -(4(2I+1)+36), so:

4(2I+1)+36

We multiply parentheses

8I+4+36

We add all the numbers together, and all the variables

8I+40

Back to the equation:

-(8I+40)



We get rid of parentheses

-8I-40+85=0

We add all the numbers together, and all the variables

-8I+45=0

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

-8I=-45

I=-45/-8

I=5+5/8

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