9/4k+9/5=2+3/7k

Solution for 9/4k+9/5=2+3/7k equation:

9/4k+9/5=2+3/7k

We move all terms to the left:

9/4k+9/5-(2+3/7k)=0


Domain of the equation: 4k!=0

k!=0/4

k!=0

k∈R



Domain of the equation: 7k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

9/4k-(3/7k+2)+9/5=0

We get rid of parentheses

9/4k-3/7k-2+9/5=0

We calculate fractions


1764k^2/700k^2+1575k/700k^2+(-300k)/700k^2-2=0

We multiply all the terms by the denominator


1764k^2+1575k+(-300k)-2*700k^2=0

Wy multiply elements

1764k^2-1400k^2+1575k+(-300k)=0

We get rid of parentheses

1764k^2-1400k^2+1575k-300k=0

We add all the numbers together, and all the variables

364k^2+1275k=0

a = 364; b = 1275; c = 0;
Δ = b2-4ac
Δ = 12752-4·364·0
Δ = 1625625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1625625}=1275$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1275)-1275}{2*364}=\frac{-2550}{728}
=-3+183/364
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1275)+1275}{2*364}=\frac{0}{728}
=0
$