1/5y+3y=2y+42

Solution for 1/5y+3y=2y+42 equation:

1/5y+3y=2y+42

We move all terms to the left:

1/5y+3y-(2y+42)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We add all the numbers together, and all the variables

3y+1/5y-(2y+42)=0

We get rid of parentheses

3y+1/5y-2y-42=0

We multiply all the terms by the denominator


3y*5y-2y*5y-42*5y+1=0

Wy multiply elements

15y^2-10y^2-210y+1=0

We add all the numbers together, and all the variables

5y^2-210y+1=0

a = 5; b = -210; c = +1;
Δ = b2-4ac
Δ = -2102-4·5·1
Δ = 44080
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44080}=\sqrt{16*2755}=\sqrt{16}*\sqrt{2755}=4\sqrt{2755}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-210)-4\sqrt{2755}}{2*5}=\frac{210-4\sqrt{2755}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-210)+4\sqrt{2755}}{2*5}=\frac{210+4\sqrt{2755}}{10}
$