(5/r)+(2/7r)=1

Solution for (5/r)+(2/7r)=1 equation:

(5/r)+(2/7r)=1

We move all terms to the left:

(5/r)+(2/7r)-(1)=0


Domain of the equation: r)!=0

r!=0/1

r!=0

r∈R



Domain of the equation: 7r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

(+5/r)+(+2/7r)-1=0

We get rid of parentheses

5/r+2/7r-1=0

We calculate fractions


35r/7r^2+2r/7r^2-1=0

We multiply all the terms by the denominator


35r+2r-1*7r^2=0

We add all the numbers together, and all the variables

37r-1*7r^2=0

Wy multiply elements

-7r^2+37r=0

a = -7; b = 37; c = 0;
Δ = b2-4ac
Δ = 372-4·(-7)·0
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-37}{2*-7}=\frac{-74}{-14}
=5+2/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+37}{2*-7}=\frac{0}{-14}
=0
$