7/8(-32-48)+36=2/3(-33x-18)-10x

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Solution for 7/8(-32-48)+36=2/3(-33x-18)-10x equation: 



7/8(-32-48)+36=2/3(-33x-18)-10x

We move all terms to the left:

7/8(-32-48)+36-(2/3(-33x-18)-10x)=0



Domain of the equation: 3(-33x-18)-10x)!=0

x∈R



We add all the numbers together, and all the variables

-(2/3(-33x-18)-10x)+36+7/8(-80)=0

We calculate fractions


(-210x^2)/(-99x-10x)*8(-54)+()/(-99x-10x)*8(-54)+36=0

We add all the numbers together, and all the variables

(-210x^2)/(-109x)*8(-54)+()/(-109x)*8(-54)+36=0

We multiply all the terms by the denominator


(-210x^2)+36*(-109x)*8(-54)+()=0

We add all the numbers together, and all the variables

(-210x^2)+36*(-109x)*8(-54)=0

We get rid of parentheses

-210x^2+36*(-109x)*8(-54)=0

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