-1/5u-7/3=1/3u-4

Solution for -1/5u-7/3=1/3u-4 equation:

-1/5u-7/3=1/3u-4

We move all terms to the left:

-1/5u-7/3-(1/3u-4)=0


Domain of the equation: 5u!=0

u!=0/5

u!=0

u∈R



Domain of the equation: 3u-4)!=0

u∈R


We get rid of parentheses

-1/5u-1/3u+4-7/3=0

We calculate fractions


(-27u)/135u^2+(-5u)/135u^2+(-35u)/135u^2+4=0

We multiply all the terms by the denominator


(-27u)+(-5u)+(-35u)+4*135u^2=0

Wy multiply elements

540u^2+(-27u)+(-5u)+(-35u)=0

We get rid of parentheses

540u^2-27u-5u-35u=0

We add all the numbers together, and all the variables

540u^2-67u=0

a = 540; b = -67; c = 0;
Δ = b2-4ac
Δ = -672-4·540·0
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-67}{2*540}=\frac{0}{1080}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+67}{2*540}=\frac{134}{1080}
=67/540
$