Question

solve the equation 1 . 1/3 (2x-1) -1/4 (2x+1) 1/12 (2-x)​

Answer 1

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

We move all terms to the left:

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

Domain of the equation: 3(2x-1)!=0

Domain of the equation: 12(2-x))!=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions

-2x+(12x(-)/(3(2x-1)*12(-1x+2)))+(-3x2/(3(2x-1)*12(-1x+2)))-1=0

We calculate terms in parentheses: +(12x(-)/(3(2x-1)*12(-1x+2))), so:

12x(-)/(3(2x-1)*12(-1x+2))

We add all the numbers together, and all the variables

12x0/(3(2x-1)*12(-1x+2))

We multiply all the terms by the denominator

12x0

We add all the numbers together, and all the variables

12x

Back to the equation:

+(12x)

We calculate terms in parentheses: +(-3x2/(3(2x-1)*12(-1x+2))), so:

-3x2/(3(2x-1)*12(-1x+2))

We multiply all the terms by the denominator

-3x2

We add all the numbers together, and all the variables

-3x^2

Back to the equation:

+(-3x^2)

We add all the numbers together, and all the variables

(-3x^2)+10x-1=0

We get rid of parentheses

-3x^2+10x-1=0

a = -3; b = 10; c = -1;

Δ = b2-4ac

Δ = 102-4·(-3)·(-1)

Δ = 88

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

x1=−b−Δ√2ax2=−b+Δ√2a

The end solution:

Δ−−√=88−−√=4∗22−−−−−√=4√∗22−−√=222−−√

x1=−b−Δ√2a=−(10)−222√2∗−3=−10−222√−6

x2=−b+Δ√2a=−(10)+222√2∗−3=−10+222√−6