(2/x-7)-(5/6x-42)=16

Solution for (2/x-7)-(5/6x-42)=16 equation:

(2/x-7)-(5/6x-42)=16

We move all terms to the left:

(2/x-7)-(5/6x-42)-(16)=0


Domain of the equation: x-7)!=0

x∈R



Domain of the equation: 6x-42)!=0

x∈R


We get rid of parentheses

2/x-5/6x-7+42-16=0

We calculate fractions


12x/6x^2+(-5x)/6x^2-7+42-16=0

We add all the numbers together, and all the variables

12x/6x^2+(-5x)/6x^2+19=0

We multiply all the terms by the denominator


12x+(-5x)+19*6x^2=0

Wy multiply elements

114x^2+12x+(-5x)=0

We get rid of parentheses

114x^2+12x-5x=0

We add all the numbers together, and all the variables

114x^2+7x=0

a = 114; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·114·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*114}=\frac{-14}{228}
=-7/114
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*114}=\frac{0}{228}
=0
$