2k-8=1/3k-1

Solution for 2k-8=1/3k-1 equation:

2k-8=1/3k-1

We move all terms to the left:

2k-8-(1/3k-1)=0


Domain of the equation: 3k-1)!=0

k∈R


We get rid of parentheses

2k-1/3k+1-8=0

We multiply all the terms by the denominator


2k*3k+1*3k-8*3k-1=0

Wy multiply elements

6k^2+3k-24k-1=0

We add all the numbers together, and all the variables

6k^2-21k-1=0

a = 6; b = -21; c = -1;
Δ = b2-4ac
Δ = -212-4·6·(-1)
Δ = 465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{465}}{2*6}=\frac{21-\sqrt{465}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{465}}{2*6}=\frac{21+\sqrt{465}}{12}
$