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

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



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

We move all terms to the left:

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



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

We move all terms containing x to the left, all other terms to the right

2x!=2

x!=2/2

x!=1

x∈R





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

We move all terms containing x to the left, all other terms to the right

2x!=3

x!=3/2

x!=1+1/2

x∈R



We calculate fractions


((x+1)*(2x-3))/((2x-2)*(2x-3))+(6x^2-6x)/((2x-2)*(2x-3))-4=0



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

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

We multiply all the terms by the denominator


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

Back to the equation:

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





We calculate terms in parentheses: +(6x^2-6x)/((2x-2)*(2x-3)), so:

6x^2-6x)/((2x-2)*(2x-3)

We multiply all the terms by the denominator


6x^2*((2x-2)*(2x-3)-6x)

Back to the equation:

+(6x^2*((2x-2)*(2x-3)-6x))

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