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1/5x+9=3/2x+4
We move all terms to the left:
1/5x+9-(3/2x+4)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 2x+4)!=0We get rid of parentheses
x∈R
1/5x-3/2x-4+9=0
We calculate fractions
2x/10x^2+(-15x)/10x^2-4+9=0
We add all the numbers together, and all the variables
2x/10x^2+(-15x)/10x^2+5=0
We multiply all the terms by the denominator
2x+(-15x)+5*10x^2=0
Wy multiply elements
50x^2+2x+(-15x)=0
We get rid of parentheses
50x^2+2x-15x=0
We add all the numbers together, and all the variables
50x^2-13x=0
a = 50; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·50·0
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*50}=\frac{0}{100} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*50}=\frac{26}{100} =13/50 $
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