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2/3x+10=5/9x-5
We move all terms to the left:
2/3x+10-(5/9x-5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 9x-5)!=0We get rid of parentheses
x∈R
2/3x-5/9x+5+10=0
We calculate fractions
18x/27x^2+(-15x)/27x^2+5+10=0
We add all the numbers together, and all the variables
18x/27x^2+(-15x)/27x^2+15=0
We multiply all the terms by the denominator
18x+(-15x)+15*27x^2=0
Wy multiply elements
405x^2+18x+(-15x)=0
We get rid of parentheses
405x^2+18x-15x=0
We add all the numbers together, and all the variables
405x^2+3x=0
a = 405; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·405·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*405}=\frac{-6}{810} =-1/135 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*405}=\frac{0}{810} =0 $
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