(2/7x)+(5/49x)=-19

Solution for (2/7x)+(5/49x)=-19 equation:

(2/7x)+(5/49x)=-19

We move all terms to the left:

(2/7x)+(5/49x)-(-19)=0


Domain of the equation: 7x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 49x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/7x)+(+5/49x)-(-19)=0

We add all the numbers together, and all the variables

(+2/7x)+(+5/49x)+19=0

We get rid of parentheses

2/7x+5/49x+19=0

We calculate fractions


98x/343x^2+35x/343x^2+19=0

We multiply all the terms by the denominator


98x+35x+19*343x^2=0

We add all the numbers together, and all the variables

133x+19*343x^2=0

Wy multiply elements

6517x^2+133x=0

a = 6517; b = 133; c = 0;
Δ = b2-4ac
Δ = 1332-4·6517·0
Δ = 17689
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17689}=133$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(133)-133}{2*6517}=\frac{-266}{13034}
=-1/49
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(133)+133}{2*6517}=\frac{0}{13034}
=0
$