3/2y+y=9/2y

Solution for 3/2y+y=9/2y equation:

3/2y+y=9/2y

We move all terms to the left:

3/2y+y-(9/2y)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 2y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

3/2y+y-(+9/2y)=0

We add all the numbers together, and all the variables

y+3/2y-(+9/2y)=0

We get rid of parentheses

y+3/2y-9/2y=0

We multiply all the terms by the denominator


y*2y+3-9=0

We add all the numbers together, and all the variables

y*2y-6=0

Wy multiply elements

2y^2-6=0

a = 2; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·2·(-6)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*2}=\frac{0-4\sqrt{3}}{4}
=-\frac{4\sqrt{3}}{4}
=-\sqrt{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*2}=\frac{0+4\sqrt{3}}{4}
=\frac{4\sqrt{3}}{4}
=\sqrt{3}
$