I/5(2x-15)=1/10(4x-30)

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Solution for I/5(2x-15)=1/10(4x-30) equation: 



/5(2I-15)=1/10(4I-30)

We move all terms to the left:

/5(2I-15)-(1/10(4I-30))=0



Domain of the equation: 5(2I-15)!=0

I∈R





Domain of the equation: 10(4I-30))!=0

I∈R



We calculate fractions


(*10(4I-30)))/(5(2I-15)*10(4I-30)))+(-5I2/(5(2I-15)*10(4I-30)))=0



We calculate terms in parentheses: +(*10(4I-30)))/(5(2I-15)*10(4I-30)))+(-5I2/(5(2I-15)*10(4I-30))), so:

*10(4I-30)))/(5(2I-15)*10(4I-30)))+(-5I2/(5(2I-15)*10(4I-30))

We add all the numbers together, and all the variables

*10(4I-30)))/(5(2I-15)*10(4I-30)))+(-5I2/(5(2I-15)*10(4I

We multiply all the terms by the denominator


*10(4I-30)))-15*104I*(5(2I-30)))+(-5I2-15*104I*(5(2I

Back to the equation:

+(*10(4I-30)))-15*104I*(5(2I-30)))+(-5I2-15*104I*(5(2I)

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