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1/3x+10=3/4x+5
We move all terms to the left:
1/3x+10-(3/4x+5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 4x+5)!=0We get rid of parentheses
x∈R
1/3x-3/4x-5+10=0
We calculate fractions
4x/12x^2+(-9x)/12x^2-5+10=0
We add all the numbers together, and all the variables
4x/12x^2+(-9x)/12x^2+5=0
We multiply all the terms by the denominator
4x+(-9x)+5*12x^2=0
Wy multiply elements
60x^2+4x+(-9x)=0
We get rid of parentheses
60x^2+4x-9x=0
We add all the numbers together, and all the variables
60x^2-5x=0
a = 60; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·60·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*60}=\frac{0}{120} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*60}=\frac{10}{120} =1/12 $
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