(3/8)f+1/2=(3/8)f-18

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Solution for (3/8)f+1/2=(3/8)f-18 equation: 



(3/8)f+1/2=(3/8)f-18

We move all terms to the left:

(3/8)f+1/2-((3/8)f-18)=0



Domain of the equation: 8)f!=0

f!=0/1

f!=0

f∈R





Domain of the equation: 8)f-18)!=0

f!=0/1

f!=0

f∈R



We add all the numbers together, and all the variables

(+3/8)f-((+3/8)f-18)+1/2=0

We multiply parentheses

3f^2-((+3/8)f-18)+1/2=0

We calculate fractions


3f^2+()/8f-18)*2)+8f/8f-18)*2)=0

We calculate fractions


3f^2+(()*8f)/64f^2+(-18)*2)+8f*8f)/64f^2=0

We multiply all the terms by the denominator


3f^2*64f^2+(()*8f)+(-18)*2)+8f*8f)=0



We calculate terms in parentheses: +(()*8f), so:

()*8f



Wy multiply elements

192f^4+(()*8f)+(-18)*2)+8f*8f)=0

We get rid of parentheses

192f^4+(()*8f)-18)*2)+8f*8f=0



We calculate terms in parentheses: +(()*8f), so:

()*8f



We do not support efpression: f^4

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