(y/3)+(y/5)=(7/10)

Solution for (y/3)+(y/5)=(7/10) equation:

(y/3)+(y/5)=(7/10)

We move all terms to the left:

(y/3)+(y/5)-((7/10))=0

We add all the numbers together, and all the variables

(+y/3)+(+y/5)-((+7/10))=0

We get rid of parentheses

y/3+y/5-((+7/10))=0

We calculate fractions


50y^2/()+30y^2/()+()/()=0

We add all the numbers together, and all the variables

50y^2/()+30y^2/()+1=0

We multiply all the terms by the denominator


50y^2+30y^2+1*()=0

We add all the numbers together, and all the variables

80y^2=0

a = 80; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·80·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{160}=0$