1/6x-7=2/3x-16

Solution for 1/6x-7=2/3x-16 equation:

1/6x-7=2/3x-16

We move all terms to the left:

1/6x-7-(2/3x-16)=0


Domain of the equation: 6x!=0

x!=0/6

x!=0

x∈R



Domain of the equation: 3x-16)!=0

x∈R


We get rid of parentheses

1/6x-2/3x+16-7=0

We calculate fractions


3x/18x^2+(-12x)/18x^2+16-7=0

We add all the numbers together, and all the variables

3x/18x^2+(-12x)/18x^2+9=0

We multiply all the terms by the denominator


3x+(-12x)+9*18x^2=0

Wy multiply elements

162x^2+3x+(-12x)=0

We get rid of parentheses

162x^2+3x-12x=0

We add all the numbers together, and all the variables

162x^2-9x=0

a = 162; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·162·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*162}=\frac{0}{324}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*162}=\frac{18}{324}
=1/18
$