(2x-5)+(3x/2+10)+(4x/3+7)=128

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Solution for (2x-5)+(3x/2+10)+(4x/3+7)=128 equation: 



(2x-5)+(3x/2+10)+(4x/3+7)=128

We move all terms to the left:

(2x-5)+(3x/2+10)+(4x/3+7)-(128)=0

We get rid of parentheses

2x+3x/2+4x/3-5+10+7-128=0

We calculate fractions

There is no solution for this equation

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