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1/2x+2=2/5x+3
We move all terms to the left:
1/2x+2-(2/5x+3)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 5x+3)!=0We get rid of parentheses
x∈R
1/2x-2/5x-3+2=0
We calculate fractions
5x/10x^2+(-4x)/10x^2-3+2=0
We add all the numbers together, and all the variables
5x/10x^2+(-4x)/10x^2-1=0
We multiply all the terms by the denominator
5x+(-4x)-1*10x^2=0
Wy multiply elements
-10x^2+5x+(-4x)=0
We get rid of parentheses
-10x^2+5x-4x=0
We add all the numbers together, and all the variables
-10x^2+x=0
a = -10; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-10)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-10}=\frac{-2}{-20} =1/10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-10}=\frac{0}{-20} =0 $
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