1/3x-4=1/8x+6

Solution for 1/3x-4=1/8x+6 equation:

1/3x-4=1/8x+6

We move all terms to the left:

1/3x-4-(1/8x+6)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 8x+6)!=0

x∈R


We get rid of parentheses

1/3x-1/8x-6-4=0

We calculate fractions


8x/24x^2+(-3x)/24x^2-6-4=0

We add all the numbers together, and all the variables

8x/24x^2+(-3x)/24x^2-10=0

We multiply all the terms by the denominator


8x+(-3x)-10*24x^2=0

Wy multiply elements

-240x^2+8x+(-3x)=0

We get rid of parentheses

-240x^2+8x-3x=0

We add all the numbers together, and all the variables

-240x^2+5x=0

a = -240; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-240)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-240}=\frac{-10}{-480}
=1/48
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-240}=\frac{0}{-480}
=0
$