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4/x+3=16/5x+11
We move all terms to the left:
4/x+3-(16/5x+11)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: 5x+11)!=0We get rid of parentheses
x∈R
4/x-16/5x-11+3=0
We calculate fractions
20x/5x^2+(-16x)/5x^2-11+3=0
We add all the numbers together, and all the variables
20x/5x^2+(-16x)/5x^2-8=0
We multiply all the terms by the denominator
20x+(-16x)-8*5x^2=0
Wy multiply elements
-40x^2+20x+(-16x)=0
We get rid of parentheses
-40x^2+20x-16x=0
We add all the numbers together, and all the variables
-40x^2+4x=0
a = -40; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-40)·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-40}=\frac{-8}{-80} =1/10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-40}=\frac{0}{-80} =0 $
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