4/3y+1=1y+6

Solution for 4/3y+1=1y+6 equation:

4/3y+1=1y+6

We move all terms to the left:

4/3y+1-(1y+6)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

4/3y-(y+6)+1=0

We get rid of parentheses

4/3y-y-6+1=0

We multiply all the terms by the denominator


-y*3y-6*3y+1*3y+4=0

Wy multiply elements

-3y^2-18y+3y+4=0

We add all the numbers together, and all the variables

-3y^2-15y+4=0

a = -3; b = -15; c = +4;
Δ = b2-4ac
Δ = -152-4·(-3)·4
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{273}}{2*-3}=\frac{15-\sqrt{273}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{273}}{2*-3}=\frac{15+\sqrt{273}}{-6}
$