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3+4/5x=9/10x
We move all terms to the left:
3+4/5x-(9/10x)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
4/5x-(+9/10x)+3=0
We get rid of parentheses
4/5x-9/10x+3=0
We calculate fractions
40x/50x^2+(-45x)/50x^2+3=0
We multiply all the terms by the denominator
40x+(-45x)+3*50x^2=0
Wy multiply elements
150x^2+40x+(-45x)=0
We get rid of parentheses
150x^2+40x-45x=0
We add all the numbers together, and all the variables
150x^2-5x=0
a = 150; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·150·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*150}=\frac{0}{300} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*150}=\frac{10}{300} =1/30 $
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