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1/7x+8=1/3x
We move all terms to the left:
1/7x+8-(1/3x)=0
Domain of the equation: 7x!=0
x!=0/7
x!=0
x∈R
Domain of the equation: 3x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/7x-(+1/3x)+8=0
We get rid of parentheses
1/7x-1/3x+8=0
We calculate fractions
3x/21x^2+(-7x)/21x^2+8=0
We multiply all the terms by the denominator
3x+(-7x)+8*21x^2=0
Wy multiply elements
168x^2+3x+(-7x)=0
We get rid of parentheses
168x^2+3x-7x=0
We add all the numbers together, and all the variables
168x^2-4x=0
a = 168; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·168·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*168}=\frac{0}{336} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*168}=\frac{8}{336} =1/42 $
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