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(4)/(5)y+8=(7)/(8)y+9
We move all terms to the left:
(4)/(5)y+8-((7)/(8)y+9)=0
Domain of the equation: 5y!=0
y!=0/5
y!=0
y∈R
Domain of the equation: 8y+9)!=0We get rid of parentheses
y∈R
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 $
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