1/3(27x-18)=-1/2(-16x+20)

Solution for 1/3(27x-18)=-1/2(-16x+20) equation:

1/3(27x-18)=-1/2(-16x+20)

We move all terms to the left:

1/3(27x-18)-(-1/2(-16x+20))=0


Domain of the equation: 3(27x-18)!=0

x∈R



Domain of the equation: 2(-16x+20))!=0

x∈R


We calculate fractions


(2x(-)/(3(27x-18)*2(-16x+20)))+(-(-3x2)/(3(27x-18)*2(-16x+20)))=0


We calculate terms in parentheses: +(2x(-)/(3(27x-18)*2(-16x+20))), so:

2x(-)/(3(27x-18)*2(-16x+20))

We add all the numbers together, and all the variables

2x0/(3(27x-18)*2(-16x+20))

We multiply all the terms by the denominator


2x0

We add all the numbers together, and all the variables

2x

Back to the equation:

+(2x)



We calculate terms in parentheses: +(-(-3x2)/(3(27x-18)*2(-16x+20))), so:

-(-3x2)/(3(27x-18)*2(-16x+20))

We add all the numbers together, and all the variables

-(-3x^2)/(3(27x-18)*2(-16x+20))

We multiply all the terms by the denominator


-(-3x^2)

We get rid of parentheses

3x^2

Back to the equation:

+(3x^2)


determiningTheFunctionDomain
3x^2+2x=0

a = 3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·3·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*3}=\frac{0}{6}
=0
$