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1/4x+3=1/2x+5
We move all terms to the left:
1/4x+3-(1/2x+5)=0
Domain of the equation: 4x!=0
x!=0/4
x!=0
x∈R
Domain of the equation: 2x+5)!=0We get rid of parentheses
x∈R
1/4x-1/2x-5+3=0
We calculate fractions
2x/8x^2+(-4x)/8x^2-5+3=0
We add all the numbers together, and all the variables
2x/8x^2+(-4x)/8x^2-2=0
We multiply all the terms by the denominator
2x+(-4x)-2*8x^2=0
Wy multiply elements
-16x^2+2x+(-4x)=0
We get rid of parentheses
-16x^2+2x-4x=0
We add all the numbers together, and all the variables
-16x^2-2x=0
a = -16; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-16)·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-16}=\frac{0}{-32} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-16}=\frac{4}{-32} =-1/8 $
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