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2/3x+3=-4/2x+5
We move all terms to the left:
2/3x+3-(-4/2x+5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x+5)!=0We get rid of parentheses
x∈R
2/3x+4/2x-5+3=0
We calculate fractions
4x/6x^2+12x/6x^2-5+3=0
We add all the numbers together, and all the variables
4x/6x^2+12x/6x^2-2=0
We multiply all the terms by the denominator
4x+12x-2*6x^2=0
We add all the numbers together, and all the variables
16x-2*6x^2=0
Wy multiply elements
-12x^2+16x=0
a = -12; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-12)·0
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-12}=\frac{-32}{-24} =1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-12}=\frac{0}{-24} =0 $
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