1/2(2x-4)=1/3(-3x-12)

Solution for 1/2(2x-4)=1/3(-3x-12) equation:

1/2(2x-4)=1/3(-3x-12)

We move all terms to the left:

1/2(2x-4)-(1/3(-3x-12))=0


Domain of the equation: 2(2x-4)!=0

x∈R



Domain of the equation: 3(-3x-12))!=0

x∈R


We calculate fractions


(3x(-)/(2(2x-4)*3(-3x-12)))+(-2x2/(2(2x-4)*3(-3x-12)))=0


We calculate terms in parentheses: +(3x(-)/(2(2x-4)*3(-3x-12))), so:

3x(-)/(2(2x-4)*3(-3x-12))

We add all the numbers together, and all the variables

3x0/(2(2x-4)*3(-3x-12))

We multiply all the terms by the denominator


3x0

We add all the numbers together, and all the variables

3x

Back to the equation:

+(3x)



We calculate terms in parentheses: +(-2x2/(2(2x-4)*3(-3x-12))), so:

-2x2/(2(2x-4)*3(-3x-12))

We multiply all the terms by the denominator


-2x2

We add all the numbers together, and all the variables

-2x^2

Back to the equation:

+(-2x^2)


We get rid of parentheses

-2x^2+3x=0

a = -2; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-2)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-2}=\frac{0}{-4}
=0
$