(2y-1)=(4/3y+8)

Solution for (2y-1)=(4/3y+8) equation:

(2y-1)=(4/3y+8)

We move all terms to the left:

(2y-1)-((4/3y+8))=0


Domain of the equation: 3y+8))!=0

y∈R


We get rid of parentheses

2y-((4/3y+8))-1=0

We multiply all the terms by the denominator


2y*3y-1*3y+8))-((4+8))=0

We add all the numbers together, and all the variables

2y*3y-1*3y+8))-(12)=0

We add all the numbers together, and all the variables

2y*3y-1*3y=0

Wy multiply elements

6y^2-3y=0

a = 6; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·6·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*6}=\frac{0}{12}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*6}=\frac{6}{12}
=1/2
$