(4/2x+3)+(17/5x-3)=3

Solution for (4/2x+3)+(17/5x-3)=3 equation:

(4/2x+3)+(17/5x-3)=3

We move all terms to the left:

(4/2x+3)+(17/5x-3)-(3)=0


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

x∈R



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

x∈R


We get rid of parentheses

4/2x+17/5x+3-3-3=0

We calculate fractions


20x/10x^2+34x/10x^2+3-3-3=0

We add all the numbers together, and all the variables

20x/10x^2+34x/10x^2-3=0

We multiply all the terms by the denominator


20x+34x-3*10x^2=0

We add all the numbers together, and all the variables

54x-3*10x^2=0

Wy multiply elements

-30x^2+54x=0

a = -30; b = 54; c = 0;
Δ = b2-4ac
Δ = 542-4·(-30)·0
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-54}{2*-30}=\frac{-108}{-60}
=1+4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+54}{2*-30}=\frac{0}{-60}
=0
$