3(100d+3)=9(35d+100)+99

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Solution for 3(100d+3)=9(35d+100)+99 equation: 



3(100d+3)=9(35d+100)+99

We move all terms to the left:

3(100d+3)-(9(35d+100)+99)=0

We multiply parentheses

300d-(9(35d+100)+99)+9=0



We calculate terms in parentheses: -(9(35d+100)+99), so:

9(35d+100)+99

We multiply parentheses

315d+900+99

We add all the numbers together, and all the variables

315d+999

Back to the equation:

-(315d+999)



We get rid of parentheses

300d-315d-999+9=0

We add all the numbers together, and all the variables

-15d-990=0

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

-15d=990

d=990/-15

d=-66

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