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1/5x+3x=42
We move all terms to the left:
1/5x+3x-(42)=0
Domain of the equation: 5x!=0We add all the numbers together, and all the variables
x!=0/5
x!=0
x∈R
3x+1/5x-42=0
We multiply all the terms by the denominator
3x*5x-42*5x+1=0
Wy multiply elements
15x^2-210x+1=0
a = 15; b = -210; c = +1;
Δ = b2-4ac
Δ = -2102-4·15·1
Δ = 44040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44040}=\sqrt{4*11010}=\sqrt{4}*\sqrt{11010}=2\sqrt{11010}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-210)-2\sqrt{11010}}{2*15}=\frac{210-2\sqrt{11010}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-210)+2\sqrt{11010}}{2*15}=\frac{210+2\sqrt{11010}}{30} $
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