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2/3x-4=5/2x+3
We move all terms to the left:
2/3x-4-(5/2x+3)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x+3)!=0We get rid of parentheses
x∈R
2/3x-5/2x-3-4=0
We calculate fractions
4x/6x^2+(-15x)/6x^2-3-4=0
We add all the numbers together, and all the variables
4x/6x^2+(-15x)/6x^2-7=0
We multiply all the terms by the denominator
4x+(-15x)-7*6x^2=0
Wy multiply elements
-42x^2+4x+(-15x)=0
We get rid of parentheses
-42x^2+4x-15x=0
We add all the numbers together, and all the variables
-42x^2-11x=0
a = -42; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·(-42)·0
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*-42}=\frac{0}{-84} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*-42}=\frac{22}{-84} =-11/42 $
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