1/2y+3/4y+1=111/4

Solution for 1/2y+3/4y+1=111/4 equation:

1/2y+3/4y+1=111/4

We move all terms to the left:

1/2y+3/4y+1-(111/4)=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 add all the numbers together, and all the variables

1/2y+3/4y+1-(+111/4)=0

We get rid of parentheses

1/2y+3/4y+1-111/4=0

We calculate fractions


64y/128y^2+6y/128y^2+(-222y)/128y^2+1=0

We multiply all the terms by the denominator


64y+6y+(-222y)+1*128y^2=0

We add all the numbers together, and all the variables

70y+(-222y)+1*128y^2=0

Wy multiply elements

128y^2+70y+(-222y)=0

We get rid of parentheses

128y^2+70y-222y=0

We add all the numbers together, and all the variables

128y^2-152y=0

a = 128; b = -152; c = 0;
Δ = b2-4ac
Δ = -1522-4·128·0
Δ = 23104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{23104}=152$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-152)-152}{2*128}=\frac{0}{256}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-152)+152}{2*128}=\frac{304}{256}
=1+3/16
$