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2/3x+1=3/4x-3
We move all terms to the left:
2/3x+1-(3/4x-3)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 4x-3)!=0We get rid of parentheses
x∈R
2/3x-3/4x+3+1=0
We calculate fractions
8x/12x^2+(-9x)/12x^2+3+1=0
We add all the numbers together, and all the variables
8x/12x^2+(-9x)/12x^2+4=0
We multiply all the terms by the denominator
8x+(-9x)+4*12x^2=0
Wy multiply elements
48x^2+8x+(-9x)=0
We get rid of parentheses
48x^2+8x-9x=0
We add all the numbers together, and all the variables
48x^2-1x=0
a = 48; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·48·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*48}=\frac{0}{96} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*48}=\frac{2}{96} =1/48 $
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