7/5y+1=19/5,y=3

Solution for 7/5y+1=19/5,y=3 equation:

7/5y+1=19/5.y=3

We move all terms to the left:

7/5y+1-(19/5.y)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R



Domain of the equation: 5.y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

7/5y-(+19/5.y)+1=0

We get rid of parentheses

7/5y-19/5.y+1=0

We calculate fractions


35y/25y^2+(-95y)/25y^2+1=0

We multiply all the terms by the denominator


35y+(-95y)+1*25y^2=0

Wy multiply elements

25y^2+35y+(-95y)=0

We get rid of parentheses

25y^2+35y-95y=0

We add all the numbers together, and all the variables

25y^2-60y=0

a = 25; b = -60; c = 0;
Δ = b2-4ac
Δ = -602-4·25·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-60}{2*25}=\frac{0}{50}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+60}{2*25}=\frac{120}{50}
=2+2/5
$