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1/6x+3/4x=12
We move all terms to the left:
1/6x+3/4x-(12)=0
Domain of the equation: 6x!=0
x!=0/6
x!=0
x∈R
Domain of the equation: 4x!=0We calculate fractions
x!=0/4
x!=0
x∈R
4x/24x^2+18x/24x^2-12=0
We multiply all the terms by the denominator
4x+18x-12*24x^2=0
We add all the numbers together, and all the variables
22x-12*24x^2=0
Wy multiply elements
-288x^2+22x=0
a = -288; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-288)·0
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-288}=\frac{-44}{-576} =11/144 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-288}=\frac{0}{-576} =0 $
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