7/6x+1/12=3+7/18x

Solution for 7/6x+1/12=3+7/18x equation:

7/6x+1/12=3+7/18x

We move all terms to the left:

7/6x+1/12-(3+7/18x)=0


Domain of the equation: 6x!=0

x!=0/6

x!=0

x∈R



Domain of the equation: 18x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

7/6x-(7/18x+3)+1/12=0

We get rid of parentheses

7/6x-7/18x-3+1/12=0

We calculate fractions


108x^2/1296x^2+1512x/1296x^2+(-504x)/1296x^2-3=0

We multiply all the terms by the denominator


108x^2+1512x+(-504x)-3*1296x^2=0

Wy multiply elements

108x^2-3888x^2+1512x+(-504x)=0

We get rid of parentheses

108x^2-3888x^2+1512x-504x=0

We add all the numbers together, and all the variables

-3780x^2+1008x=0

a = -3780; b = 1008; c = 0;
Δ = b2-4ac
Δ = 10082-4·(-3780)·0
Δ = 1016064
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1016064}=1008$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1008)-1008}{2*-3780}=\frac{-2016}{-7560}
=4/15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1008)+1008}{2*-3780}=\frac{0}{-7560}
=0
$