2/10f+3/5(f+20)=-8+2/5f

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Solution for 2/10f+3/5(f+20)=-8+2/5f equation: 



2/10f+3/5(f+20)=-8+2/5f

We move all terms to the left:

2/10f+3/5(f+20)-(-8+2/5f)=0



Domain of the equation: 10f!=0

f!=0/10

f!=0

f∈R





Domain of the equation: 5(f+20)!=0

f∈R





Domain of the equation: 5f)!=0

f!=0/1

f!=0

f∈R



We add all the numbers together, and all the variables

2/10f+3/5(f+20)-(2/5f-8)=0

We get rid of parentheses

2/10f+3/5(f+20)-2/5f+8=0

We calculate fractions

We do not support efpression: f^3(f

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