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

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



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

We move all terms to the left:

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



Domain of the equation: 8(-32-48x)!=0

x∈R





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

x∈R



We add all the numbers together, and all the variables

7/8(-48x-32)-(2/3(-51)-10x)+36=0

We calculate fractions


(-10x)+7*3()/(8(-48x-32)*3(-51)-10x))+(-16x0/(8(-48x-32)*3(-51)-10x))+36=0

We get rid of parentheses

-10x+7*3()/(8(-48x-32)*3(-51)-10x))+(-16x0/(8(-48x-32)*3(-51)-10x))+36=0

We calculate fractions

We do not support expression: x^3(

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