(7/2u)+(5/3u)=1

Solution for (7/2u)+(5/3u)=1 equation:

(7/2u)+(5/3u)=1

We move all terms to the left:

(7/2u)+(5/3u)-(1)=0


Domain of the equation: 2u)!=0

u!=0/1

u!=0

u∈R



Domain of the equation: 3u)!=0

u!=0/1

u!=0

u∈R


We add all the numbers together, and all the variables

(+7/2u)+(+5/3u)-1=0

We get rid of parentheses

7/2u+5/3u-1=0

We calculate fractions


21u/6u^2+10u/6u^2-1=0

We multiply all the terms by the denominator


21u+10u-1*6u^2=0

We add all the numbers together, and all the variables

31u-1*6u^2=0

Wy multiply elements

-6u^2+31u=0

a = -6; b = 31; c = 0;
Δ = b2-4ac
Δ = 312-4·(-6)·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-31}{2*-6}=\frac{-62}{-12}
=5+1/6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+31}{2*-6}=\frac{0}{-12}
=0
$