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1/2x+3=3/4x-13
We move all terms to the left:
1/2x+3-(3/4x-13)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 4x-13)!=0We get rid of parentheses
x∈R
1/2x-3/4x+13+3=0
We calculate fractions
4x/8x^2+(-6x)/8x^2+13+3=0
We add all the numbers together, and all the variables
4x/8x^2+(-6x)/8x^2+16=0
We multiply all the terms by the denominator
4x+(-6x)+16*8x^2=0
Wy multiply elements
128x^2+4x+(-6x)=0
We get rid of parentheses
128x^2+4x-6x=0
We add all the numbers together, and all the variables
128x^2-2x=0
a = 128; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·128·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*128}=\frac{0}{256} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*128}=\frac{4}{256} =1/64 $
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