x/9x+1-7x-18=3x+5

Solution for x/9x+1-7x-18=3x+5 equation:

x/9x+1-7x-18=3x+5

We move all terms to the left:

x/9x+1-7x-18-(3x+5)=0


Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R


We add all the numbers together, and all the variables

-7x+x/9x-(3x+5)-17=0

We get rid of parentheses

-7x+x/9x-3x-5-17=0

We multiply all the terms by the denominator


-7x*9x+x-3x*9x-5*9x-17*9x=0

We add all the numbers together, and all the variables

x-7x*9x-3x*9x-5*9x-17*9x=0

Wy multiply elements

-63x^2-27x^2+x-45x-153x=0

We add all the numbers together, and all the variables

-90x^2-197x=0

a = -90; b = -197; c = 0;
Δ = b2-4ac
Δ = -1972-4·(-90)·0
Δ = 38809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{38809}=197$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-197)-197}{2*-90}=\frac{0}{-180}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-197)+197}{2*-90}=\frac{394}{-180}
=-2+17/90
$