(1/3)x+(1/5)=(1/5)x-1

Solution for (1/3)x+(1/5)=(1/5)x-1 equation:

(1/3)x+(1/5)=(1/5)x-1

We move all terms to the left:

(1/3)x+(1/5)-((1/5)x-1)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 5)x-1)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/3)x-((+1/5)x-1)+(+1/5)=0

We multiply parentheses

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

We get rid of parentheses

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

We calculate fractions


x^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{2}=0$