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1+4/5x=2+3/4x
We move all terms to the left:
1+4/5x-(2+3/4x)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 4x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
4/5x-(3/4x+2)+1=0
We get rid of parentheses
4/5x-3/4x-2+1=0
We calculate fractions
16x/20x^2+(-15x)/20x^2-2+1=0
We add all the numbers together, and all the variables
16x/20x^2+(-15x)/20x^2-1=0
We multiply all the terms by the denominator
16x+(-15x)-1*20x^2=0
Wy multiply elements
-20x^2+16x+(-15x)=0
We get rid of parentheses
-20x^2+16x-15x=0
We add all the numbers together, and all the variables
-20x^2+x=0
a = -20; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-20)·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-20}=\frac{-2}{-40} =1/20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-20}=\frac{0}{-40} =0 $
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