(3x/4)+(x/3)-(3x/8)=(1/6)

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

(3x/4)+(x/3)-(3x/8)=(1/6)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


(-648x^2)/()+3456x^2/()+1536x^2/()+()/()=0

We add all the numbers together, and all the variables

(-648x^2)/()+3456x^2/()+1536x^2/()+1=0

We multiply all the terms by the denominator


(-648x^2)+3456x^2+1536x^2+1*()=0

We add all the numbers together, and all the variables

4992x^2+(-648x^2)=0

We get rid of parentheses

4992x^2-648x^2=0

We add all the numbers together, and all the variables

4344x^2=0

a = 4344; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·4344·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{8688}=0$