1/2k-7=1/3k-4

Solution for 1/2k-7=1/3k-4 equation:

1/2k-7=1/3k-4

We move all terms to the left:

1/2k-7-(1/3k-4)=0


Domain of the equation: 2k!=0

k!=0/2

k!=0

k∈R



Domain of the equation: 3k-4)!=0

k∈R


We get rid of parentheses

1/2k-1/3k+4-7=0

We calculate fractions


3k/6k^2+(-2k)/6k^2+4-7=0

We add all the numbers together, and all the variables

3k/6k^2+(-2k)/6k^2-3=0

We multiply all the terms by the denominator


3k+(-2k)-3*6k^2=0

Wy multiply elements

-18k^2+3k+(-2k)=0

We get rid of parentheses

-18k^2+3k-2k=0

We add all the numbers together, and all the variables

-18k^2+k=0

a = -18; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-18)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-18}=\frac{-2}{-36}
=1/18
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-18}=\frac{0}{-36}
=0
$