1/3x+8=1/9x-16

Solution for 1/3x+8=1/9x-16 equation:

1/3x+8=1/9x-16

We move all terms to the left:

1/3x+8-(1/9x-16)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 9x-16)!=0

x∈R


We get rid of parentheses

1/3x-1/9x+16+8=0

We calculate fractions


9x/27x^2+(-3x)/27x^2+16+8=0

We add all the numbers together, and all the variables

9x/27x^2+(-3x)/27x^2+24=0

We multiply all the terms by the denominator


9x+(-3x)+24*27x^2=0

Wy multiply elements

648x^2+9x+(-3x)=0

We get rid of parentheses

648x^2+9x-3x=0

We add all the numbers together, and all the variables

648x^2+6x=0

a = 648; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·648·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*648}=\frac{-12}{1296}
=-1/108
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*648}=\frac{0}{1296}
=0
$