2/9x+10=6/3x-5=

Solution for 2/9x+10=6/3x-5= equation:

2/9x+10=6/3x-5=

We move all terms to the left:

2/9x+10-(6/3x-5)=0


Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R



Domain of the equation: 3x-5)!=0

x∈R


We get rid of parentheses

2/9x-6/3x+5+10=0

We calculate fractions


6x/27x^2+(-54x)/27x^2+5+10=0

We add all the numbers together, and all the variables

6x/27x^2+(-54x)/27x^2+15=0

We multiply all the terms by the denominator


6x+(-54x)+15*27x^2=0

Wy multiply elements

405x^2+6x+(-54x)=0

We get rid of parentheses

405x^2+6x-54x=0

We add all the numbers together, and all the variables

405x^2-48x=0

a = 405; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·405·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48}{2*405}=\frac{0}{810}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*405}=\frac{96}{810}
=16/135
$