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10/5x-9=8/2x-8
We move all terms to the left:
10/5x-9-(8/2x-8)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 2x-8)!=0We get rid of parentheses
x∈R
10/5x-8/2x+8-9=0
We calculate fractions
20x/10x^2+(-40x)/10x^2+8-9=0
We add all the numbers together, and all the variables
20x/10x^2+(-40x)/10x^2-1=0
We multiply all the terms by the denominator
20x+(-40x)-1*10x^2=0
Wy multiply elements
-10x^2+20x+(-40x)=0
We get rid of parentheses
-10x^2+20x-40x=0
We add all the numbers together, and all the variables
-10x^2-20x=0
a = -10; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-10)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-10}=\frac{0}{-20} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-10}=\frac{40}{-20} =-2 $
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