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1/4x-1=1/5x+3
We move all terms to the left:
1/4x-1-(1/5x+3)=0
Domain of the equation: 4x!=0
x!=0/4
x!=0
x∈R
Domain of the equation: 5x+3)!=0We get rid of parentheses
x∈R
1/4x-1/5x-3-1=0
We calculate fractions
5x/20x^2+(-4x)/20x^2-3-1=0
We add all the numbers together, and all the variables
5x/20x^2+(-4x)/20x^2-4=0
We multiply all the terms by the denominator
5x+(-4x)-4*20x^2=0
Wy multiply elements
-80x^2+5x+(-4x)=0
We get rid of parentheses
-80x^2+5x-4x=0
We add all the numbers together, and all the variables
-80x^2+x=0
a = -80; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-80)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-80}=\frac{-2}{-160} =1/80 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-80}=\frac{0}{-160} =0 $
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