(3/8)f+1/2=(6/16)f-18

Solution for (3/8)f+1/2=(6/16)f-18 equation:

(3/8)f+1/2=(6/16)f-18

We move all terms to the left:

(3/8)f+1/2-((6/16)f-18)=0


Domain of the equation: 8)f!=0

f!=0/1

f!=0

f∈R



Domain of the equation: 16)f-18)!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

(+3/8)f-((+6/16)f-18)+1/2=0

We multiply parentheses

3f^2-((+6/16)f-18)+1/2=0

We calculate fractions


3f^2+()/16f-18)*2)+16f/16f-18)*2)=0

We calculate fractions


3f^2+(()*16f)/256f^2+(-18)*2)+16f*16f)/256f^2=0

We multiply all the terms by the denominator


3f^2*256f^2+(()*16f)+(-18)*2)+16f*16f)=0


We calculate terms in parentheses: +(()*16f), so:

()*16f


Wy multiply elements

768f^4+(()*16f)+(-18)*2)+16f*16f)=0

We get rid of parentheses

768f^4+(()*16f)-18)*2)+16f*16f=0


We calculate terms in parentheses: +(()*16f), so:

()*16f


We do not support efpression: f^4