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1/2x-5=1/3x+1
We move all terms to the left:
1/2x-5-(1/3x+1)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 3x+1)!=0We get rid of parentheses
x∈R
1/2x-1/3x-1-5=0
We calculate fractions
3x/6x^2+(-2x)/6x^2-1-5=0
We add all the numbers together, and all the variables
3x/6x^2+(-2x)/6x^2-6=0
We multiply all the terms by the denominator
3x+(-2x)-6*6x^2=0
Wy multiply elements
-36x^2+3x+(-2x)=0
We get rid of parentheses
-36x^2+3x-2x=0
We add all the numbers together, and all the variables
-36x^2+x=0
a = -36; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-36)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-36}=\frac{-2}{-72} =1/36 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-36}=\frac{0}{-72} =0 $
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