2y+3/4y+y=15/4y

Solution for 2y+3/4y+y=15/4y equation:

2y+3/4y+y=15/4y

We move all terms to the left:

2y+3/4y+y-(15/4y)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 4y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

2y+3/4y+y-(+15/4y)=0

We add all the numbers together, and all the variables

3y+3/4y-(+15/4y)=0

We get rid of parentheses

3y+3/4y-15/4y=0

We multiply all the terms by the denominator


3y*4y+3-15=0

We add all the numbers together, and all the variables

3y*4y-12=0

Wy multiply elements

12y^2-12=0

a = 12; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·12·(-12)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*12}=\frac{-24}{24}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*12}=\frac{24}{24}
=1
$