3/2y+3/4y=3

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

3/2y+3/4y=3

We move all terms to the left:

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


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We calculate fractions


12y/8y^2+6y/8y^2-3=0

We multiply all the terms by the denominator


12y+6y-3*8y^2=0

We add all the numbers together, and all the variables

18y-3*8y^2=0

Wy multiply elements

-24y^2+18y=0

a = -24; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-24)·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-24}=\frac{-36}{-48}
=3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-24}=\frac{0}{-48}
=0
$