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4/3x+2=5/4x
We move all terms to the left:
4/3x+2-(5/4x)=0
Domain of the equation: 3x!=0
x!=0/3
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
4/3x-(+5/4x)+2=0
We get rid of parentheses
4/3x-5/4x+2=0
We calculate fractions
16x/12x^2+(-15x)/12x^2+2=0
We multiply all the terms by the denominator
16x+(-15x)+2*12x^2=0
Wy multiply elements
24x^2+16x+(-15x)=0
We get rid of parentheses
24x^2+16x-15x=0
We add all the numbers together, and all the variables
24x^2+x=0
a = 24; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·24·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*24}=\frac{-2}{48} =-1/24 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*24}=\frac{0}{48} =0 $
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