7x/3x-8-4=1/9x

Solution for 7x/3x-8-4=1/9x equation:

7x/3x-8-4=1/9x

We move all terms to the left:

7x/3x-8-4-(1/9x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 9x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

7x/3x-(+1/9x)-8-4=0

We add all the numbers together, and all the variables

7x/3x-(+1/9x)-12=0

We get rid of parentheses

7x/3x-1/9x-12=0

We calculate fractions


63x^2/27x^2+(-3x)/27x^2-12=0

We multiply all the terms by the denominator


63x^2+(-3x)-12*27x^2=0

Wy multiply elements

63x^2-324x^2+(-3x)=0

We get rid of parentheses

63x^2-324x^2-3x=0

We add all the numbers together, and all the variables

-261x^2-3x=0

a = -261; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-261)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-261}=\frac{0}{-522}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-261}=\frac{6}{-522}
=-1/87
$