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

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
$