2-2/3k=1/8k+9

Solution for 2-2/3k=1/8k+9 equation:

2-2/3k=1/8k+9

We move all terms to the left:

2-2/3k-(1/8k+9)=0


Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R



Domain of the equation: 8k+9)!=0

k∈R


We get rid of parentheses

-2/3k-1/8k-9+2=0

We calculate fractions


(-16k)/24k^2+(-3k)/24k^2-9+2=0

We add all the numbers together, and all the variables

(-16k)/24k^2+(-3k)/24k^2-7=0

We multiply all the terms by the denominator


(-16k)+(-3k)-7*24k^2=0

Wy multiply elements

-168k^2+(-16k)+(-3k)=0

We get rid of parentheses

-168k^2-16k-3k=0

We add all the numbers together, and all the variables

-168k^2-19k=0

a = -168; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·(-168)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*-168}=\frac{0}{-336}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*-168}=\frac{38}{-336}
=-19/168
$