1/4x-3=1/8x+1

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Solution for 1/4x-3=1/8x+1 equation:



1/4x-3=1/8x+1
We move all terms to the left:
1/4x-3-(1/8x+1)=0
Domain of the equation: 4x!=0
x!=0/4
x!=0
x∈R
Domain of the equation: 8x+1)!=0
x∈R
We get rid of parentheses
1/4x-1/8x-1-3=0
We calculate fractions
8x/32x^2+(-4x)/32x^2-1-3=0
We add all the numbers together, and all the variables
8x/32x^2+(-4x)/32x^2-4=0
We multiply all the terms by the denominator
8x+(-4x)-4*32x^2=0
Wy multiply elements
-128x^2+8x+(-4x)=0
We get rid of parentheses
-128x^2+8x-4x=0
We add all the numbers together, and all the variables
-128x^2+4x=0
a = -128; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-128)·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*-128}=\frac{-8}{-256} =1/32 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-128}=\frac{0}{-256} =0 $

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