(2x-5)/7x+(x+1)/3x=-4

Solution for (2x-5)/7x+(x+1)/3x=-4 equation:

(2x-5)/7x+(x+1)/3x=-4

We move all terms to the left:

(2x-5)/7x+(x+1)/3x-(-4)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

(2x-5)/7x+(x+1)/3x+4=0

We calculate fractions


(6x^2-15x)/21x^2+(7x^2+7x)/21x^2+4=0

We multiply all the terms by the denominator


(6x^2-15x)+(7x^2+7x)+4*21x^2=0

Wy multiply elements

84x^2+(6x^2-15x)+(7x^2+7x)=0

We get rid of parentheses

84x^2+6x^2+7x^2-15x+7x=0

We add all the numbers together, and all the variables

97x^2-8x=0

a = 97; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·97·0
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*97}=\frac{0}{194}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*97}=\frac{16}{194}
=8/97
$