x-5/7x-35=x+5/3x+15

Solution for x-5/7x-35=x+5/3x+15 equation:

x-5/7x-35=x+5/3x+15

We move all terms to the left:

x-5/7x-35-(x+5/3x+15)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



Domain of the equation: 3x+15)!=0

x∈R


We get rid of parentheses

x-5/7x-x-5/3x-15-35=0

We calculate fractions


x-x+(-15x)/21x^2+(-35x)/21x^2-15-35=0

We add all the numbers together, and all the variables

(-15x)/21x^2+(-35x)/21x^2-50=0

We multiply all the terms by the denominator


(-15x)+(-35x)-50*21x^2=0

Wy multiply elements

-1050x^2+(-15x)+(-35x)=0

We get rid of parentheses

-1050x^2-15x-35x=0

We add all the numbers together, and all the variables

-1050x^2-50x=0

a = -1050; b = -50; c = 0;
Δ = b2-4ac
Δ = -502-4·(-1050)·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-50}{2*-1050}=\frac{0}{-2100}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+50}{2*-1050}=\frac{100}{-2100}
=-1/21
$