(3/4)t+(1/8)=(3/4)t+6

Solution for (3/4)t+(1/8)=(3/4)t+6 equation:

(3/4)t+(1/8)=(3/4)t+6

We move all terms to the left:

(3/4)t+(1/8)-((3/4)t+6)=0


Domain of the equation: 4)t!=0

t!=0/1

t!=0

t∈R



Domain of the equation: 4)t+6)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+3/4)t-((+3/4)t+6)+(+1/8)=0

We multiply parentheses

3t^2-((+3/4)t+6)+(+1/8)=0

We get rid of parentheses

3t^2-((+3/4)t+6)+1/8=0

We calculate fractions


3t^2+()/4t+6)*8)+4t/4t+6)*8)=0

We calculate fractions


3t^2+(()*4t)/16t^2+6*8)+4t*4t)/16t^2=0

We multiply all the terms by the denominator


3t^2*16t^2+(()*4t)+6*8)+4t*4t)=0


We calculate terms in parentheses: +(()*4t), so:

()*4t


Wy multiply elements

48t^4+192t^2+(()*4t)=0


We calculate terms in parentheses: +(()*4t), so:

()*4t


We do not support etpression: t^4