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

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



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



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

x∈R





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

x∈R



We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

8x+1*15x^2=0

Wy multiply elements

15x^2+8x=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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


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


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*15}=\frac{0}{30}
=0
$

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