I0x-(2x+9)=-3(15-4x)

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Solution for I0x-(2x+9)=-3(15-4x) equation: 



0I-(2I+9)=-3(15-4I)

We move all terms to the left:

0I-(2I+9)-(-3(15-4I))=0

We add all the numbers together, and all the variables

0I-(2I+9)-(-3(-4I+15))=0

We add all the numbers together, and all the variables

I-(2I+9)-(-3(-4I+15))=0

We get rid of parentheses

I-2I-(-3(-4I+15))-9=0



We calculate terms in parentheses: -(-3(-4I+15)), so:

-3(-4I+15)

We multiply parentheses

12I-45

Back to the equation:

-(12I-45)



We add all the numbers together, and all the variables

-1I-(12I-45)-9=0

We get rid of parentheses

-1I-12I+45-9=0

We add all the numbers together, and all the variables

-13I+36=0

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

-13I=-36

I=-36/-13

I=2+10/13

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