1/2(7x+48)=(1/2x-3)+(x+5)

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Solution for 1/2(7x+48)=(1/2x-3)+(x+5) equation: 



1/2(7x+48)=(1/2x-3)+(x+5)

We move all terms to the left:

1/2(7x+48)-((1/2x-3)+(x+5))=0



Domain of the equation: 2(7x+48)!=0

x∈R





Domain of the equation: 2x-3)+(x+5))!=0

x∈R



We calculate fractions


(2x-3)+x/(28x^2+192x-3)+x+(-((1*2(7x+48))/(28x^2+192x-3)+x=0

We add all the numbers together, and all the variables

x+(2x-3)+x/(28x^2+192x-3)+(-((1*2(7x+48))/(28x^2+192x-3)+x=0

We get rid of parentheses

x+2x+x/(28x^2+192x-3)+(-((1*2(7x+48))/(28x^2+192x-3)+x-3=0

We calculate fractions

We do not support expression: x^3

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