(3/2x+1)+(4/5x-1)=2

Solution for (3/2x+1)+(4/5x-1)=2 equation:

(3/2x+1)+(4/5x-1)=2

We move all terms to the left:

(3/2x+1)+(4/5x-1)-(2)=0


Domain of the equation: 2x+1)!=0

x∈R



Domain of the equation: 5x-1)!=0

x∈R


We get rid of parentheses

3/2x+4/5x+1-1-2=0

We calculate fractions


15x/10x^2+8x/10x^2+1-1-2=0

We add all the numbers together, and all the variables

15x/10x^2+8x/10x^2-2=0

We multiply all the terms by the denominator


15x+8x-2*10x^2=0

We add all the numbers together, and all the variables

23x-2*10x^2=0

Wy multiply elements

-20x^2+23x=0

a = -20; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·(-20)·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*-20}=\frac{-46}{-40}
=1+3/20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*-20}=\frac{0}{-40}
=0
$