1/5z+3=1/4z+5

Solution for 1/5z+3=1/4z+5 equation:

1/5z+3=1/4z+5

We move all terms to the left:

1/5z+3-(1/4z+5)=0


Domain of the equation: 5z!=0

z!=0/5

z!=0

z∈R



Domain of the equation: 4z+5)!=0

z∈R


We get rid of parentheses

1/5z-1/4z-5+3=0

We calculate fractions


4z/20z^2+(-5z)/20z^2-5+3=0

We add all the numbers together, and all the variables

4z/20z^2+(-5z)/20z^2-2=0

We multiply all the terms by the denominator


4z+(-5z)-2*20z^2=0

Wy multiply elements

-40z^2+4z+(-5z)=0

We get rid of parentheses

-40z^2+4z-5z=0

We add all the numbers together, and all the variables

-40z^2-1z=0

a = -40; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-40)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-40}=\frac{0}{-80}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-40}=\frac{2}{-80}
=-1/40
$