1/5x+1300=4/7x.

Solution for 1/5x+1300=4/7x. equation:

1/5x+1300=4/7x.

We move all terms to the left:

1/5x+1300-(4/7x.)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 7x.)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

1/5x-(+4/7x.)+1300=0

We get rid of parentheses

1/5x-4/7x.+1300=0

We calculate fractions


7x/35x^2+(-20x)/35x^2+1300=0

We multiply all the terms by the denominator


7x+(-20x)+1300*35x^2=0

Wy multiply elements

45500x^2+7x+(-20x)=0

We get rid of parentheses

45500x^2+7x-20x=0

We add all the numbers together, and all the variables

45500x^2-13x=0

a = 45500; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·45500·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*45500}=\frac{0}{91000}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*45500}=\frac{26}{91000}
=1/3500
$