1/5j+1/3j=2/9

Solution for 1/5j+1/3j=2/9 equation:

1/5j+1/3j=2/9

We move all terms to the left:

1/5j+1/3j-(2/9)=0


Domain of the equation: 5j!=0

j!=0/5

j!=0

j∈R



Domain of the equation: 3j!=0

j!=0/3

j!=0

j∈R


We add all the numbers together, and all the variables

1/5j+1/3j-(+2/9)=0

We get rid of parentheses

1/5j+1/3j-2/9=0

We calculate fractions


(-90j^2)/1215j^2+243j/1215j^2+405j/1215j^2=0

We multiply all the terms by the denominator


(-90j^2)+243j+405j=0

We add all the numbers together, and all the variables

(-90j^2)+648j=0

We get rid of parentheses

-90j^2+648j=0

a = -90; b = 648; c = 0;
Δ = b2-4ac
Δ = 6482-4·(-90)·0
Δ = 419904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{419904}=648$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(648)-648}{2*-90}=\frac{-1296}{-180}
=7+1/5
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(648)+648}{2*-90}=\frac{0}{-180}
=0
$