3/7x+1/9x+6=15

Solution for 3/7x+1/9x+6=15 equation:

3/7x+1/9x+6=15

We move all terms to the left:

3/7x+1/9x+6-(15)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R


We add all the numbers together, and all the variables

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

We calculate fractions


27x/63x^2+7x/63x^2-9=0

We multiply all the terms by the denominator


27x+7x-9*63x^2=0

We add all the numbers together, and all the variables

34x-9*63x^2=0

Wy multiply elements

-567x^2+34x=0

a = -567; b = 34; c = 0;
Δ = b2-4ac
Δ = 342-4·(-567)·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-34}{2*-567}=\frac{-68}{-1134}
=34/567
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+34}{2*-567}=\frac{0}{-1134}
=0
$