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(1/3x+5)+x=48
We move all terms to the left:
(1/3x+5)+x-(48)=0
Domain of the equation: 3x+5)!=0We add all the numbers together, and all the variables
x∈R
x+(1/3x+5)-48=0
We get rid of parentheses
x+1/3x+5-48=0
We multiply all the terms by the denominator
x*3x+5*3x-48*3x+1=0
Wy multiply elements
3x^2+15x-144x+1=0
We add all the numbers together, and all the variables
3x^2-129x+1=0
a = 3; b = -129; c = +1;
Δ = b2-4ac
Δ = -1292-4·3·1
Δ = 16629
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-129)-\sqrt{16629}}{2*3}=\frac{129-\sqrt{16629}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-129)+\sqrt{16629}}{2*3}=\frac{129+\sqrt{16629}}{6} $
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