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1/2x-9=1/5x+3
We move all terms to the left:
1/2x-9-(1/5x+3)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 5x+3)!=0We get rid of parentheses
x∈R
1/2x-1/5x-3-9=0
We calculate fractions
5x/10x^2+(-2x)/10x^2-3-9=0
We add all the numbers together, and all the variables
5x/10x^2+(-2x)/10x^2-12=0
We multiply all the terms by the denominator
5x+(-2x)-12*10x^2=0
Wy multiply elements
-120x^2+5x+(-2x)=0
We get rid of parentheses
-120x^2+5x-2x=0
We add all the numbers together, and all the variables
-120x^2+3x=0
a = -120; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-120)·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-120}=\frac{-6}{-240} =1/40 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-120}=\frac{0}{-240} =0 $
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