4/5y+8=7/8y+9

Solution for 4/5y+8=7/8y+9 equation:

4/5y+8=7/8y+9

We move all terms to the left:

4/5y+8-(7/8y+9)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R



Domain of the equation: 8y+9)!=0

y∈R


We get rid of parentheses

4/5y-7/8y-9+8=0

We calculate fractions


32y/40y^2+(-35y)/40y^2-9+8=0

We add all the numbers together, and all the variables

32y/40y^2+(-35y)/40y^2-1=0

We multiply all the terms by the denominator


32y+(-35y)-1*40y^2=0

Wy multiply elements

-40y^2+32y+(-35y)=0

We get rid of parentheses

-40y^2+32y-35y=0

We add all the numbers together, and all the variables

-40y^2-3y=0

a = -40; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-40)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-40}=\frac{0}{-80}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-40}=\frac{6}{-80}
=-3/40
$