2/3x-5+8/9x=19

Solution for 2/3x-5+8/9x=19 equation:

2/3x-5+8/9x=19

We move all terms to the left:

2/3x-5+8/9x-(19)=0


Domain of the equation: 3x!=0

x!=0/3

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

2/3x+8/9x-24=0

We calculate fractions


18x/27x^2+24x/27x^2-24=0

We multiply all the terms by the denominator


18x+24x-24*27x^2=0

We add all the numbers together, and all the variables

42x-24*27x^2=0

Wy multiply elements

-648x^2+42x=0

a = -648; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·(-648)·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*-648}=\frac{-84}{-1296}
=7/108
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*-648}=\frac{0}{-1296}
=0
$