(1/3a-2)+(1/5a+3)=0

Solution for (1/3a-2)+(1/5a+3)=0 equation:

(1/3a-2)+(1/5a+3)=0


Domain of the equation: 3a-2)!=0

a∈R



Domain of the equation: 5a+3)!=0

a∈R


We get rid of parentheses

1/3a+1/5a-2+3=0

We calculate fractions


5a/15a^2+3a/15a^2-2+3=0

We add all the numbers together, and all the variables

5a/15a^2+3a/15a^2+1=0

We multiply all the terms by the denominator


5a+3a+1*15a^2=0

We add all the numbers together, and all the variables

8a+1*15a^2=0

Wy multiply elements

15a^2+8a=0

a = 15; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·15·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*15}=\frac{-16}{30}
=-8/15
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*15}=\frac{0}{30}
=0
$