1/4y+7=18-1/12y

Solution for 1/4y+7=18-1/12y equation:

1/4y+7=18-1/12y

We move all terms to the left:

1/4y+7-(18-1/12y)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 12y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/4y-(-1/12y+18)+7=0

We get rid of parentheses

1/4y+1/12y-18+7=0

We calculate fractions


12y/48y^2+4y/48y^2-18+7=0

We add all the numbers together, and all the variables

12y/48y^2+4y/48y^2-11=0

We multiply all the terms by the denominator


12y+4y-11*48y^2=0

We add all the numbers together, and all the variables

16y-11*48y^2=0

Wy multiply elements

-528y^2+16y=0

a = -528; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-528)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-528}=\frac{-32}{-1056}
=1/33
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-528}=\frac{0}{-1056}
=0
$