(2/3)x+1/6=-(3/4)x+1

Solution for (2/3)x+1/6=-(3/4)x+1 equation:

(2/3)x+1/6=-(3/4)x+1

We move all terms to the left:

(2/3)x+1/6-(-(3/4)x+1)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 4)x+1)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)x-(-(+3/4)x+1)+1/6=0

We multiply parentheses

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

We calculate fractions


2x^2+()/4x+1)*6)+4x/4x+1)*6)=0

We calculate fractions


2x^2+(()*4x)/16x^2+1*6)+4x*4x)/16x^2=0

We multiply all the terms by the denominator


2x^2*16x^2+(()*4x)+1*6)+4x*4x)=0


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

()*4x


Wy multiply elements

32x^4+24x^2+(()*4x)=0


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

()*4x


We do not support expression: x^4