(2/5x)+(1/x)=35

Solution for (2/5x)+(1/x)=35 equation:

(2/5x)+(1/x)=35

We move all terms to the left:

(2/5x)+(1/x)-(35)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/5x)+(+1/x)-35=0

We get rid of parentheses

2/5x+1/x-35=0

We calculate fractions


2x/5x^2+5x/5x^2-35=0

We multiply all the terms by the denominator


2x+5x-35*5x^2=0

We add all the numbers together, and all the variables

7x-35*5x^2=0

Wy multiply elements

-175x^2+7x=0

a = -175; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-175)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-175}=\frac{-14}{-350}
=1/25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-175}=\frac{0}{-350}
=0
$