(8/7)-(5x-3)=(1/4)x-12

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Solution for (8/7)-(5x-3)=(1/4)x-12 equation: 



(8/7)-(5x-3)=(1/4)x-12

We move all terms to the left:

(8/7)-(5x-3)-((1/4)x-12)=0



Domain of the equation: 4)x-12)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

-(5x-3)-((+1/4)x-12)+(+8/7)=0

We get rid of parentheses

-5x-((+1/4)x-12)+3+8/7=0

We calculate fractions


-5x+()/4x-12)*7)+32x/4x-12)*7)+3=0

We calculate fractions


-5x+(()*4x)/16x^2+(-12)*7)+32x*4x)/16x^2=0

We multiply all the terms by the denominator


-5x*16x^2+(()*4x)+(-12)*7)+32x*4x)=0



We calculate terms in parentheses: +(()*4x), so:

()*4x



Wy multiply elements

-80x^3+(()*4x)+(-12)*7)+32x*4x)=0

We get rid of parentheses

-80x^3+(()*4x)-12)*7)+32x*4x=0



We calculate terms in parentheses: +(()*4x), so:

()*4x



We do not support expression: x^3

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