1-x+1/2=2(x+1)/5-x

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Solution for 1-x+1/2=2(x+1)/5-x equation: 



1-x+1/2=2(x+1)/5-x

We move all terms to the left:

1-x+1/2-(2(x+1)/5-x)=0



Domain of the equation: 5-x)!=0

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

-x)!=-5

x!=-5/1

x!=-5

x∈R



We add all the numbers together, and all the variables

-1x-(2(x+1)/5-x)+1+1/2=0

We calculate fractions


-1x+(-(2(x+1)*2)/(-x)*2+5)+(-x)-x)*2+5)+1+1*5)/(=0



We calculate terms in parentheses: +(-(2(x+1)*2)/(-x)*2+5), so:

-(2(x+1)*2)/(-x)*2+5

We add all the numbers together, and all the variables

-(2(x+1)*2)/(-1x)*2+5

We multiply all the terms by the denominator


-(2(x+1)*2)+5*(-1x)*2



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

2(x+1)*2

We multiply parentheses

4x+4

Back to the equation:

-(4x+4)



We multiply parentheses

-(4x+4)-10x

We get rid of parentheses

-4x-10x-4

We add all the numbers together, and all the variables

-14x-4

Back to the equation:

+(-14x-4)



We add all the numbers together, and all the variables

-1x+(-14x-4)+(-1x)-x)*2+5)+1+1*5)/(=0

We get rid of parentheses

-1x-14x-1x-x)*2+5)+1+1*5)/(-4=0

We multiply all the terms by the denominator


-1x*(-4-14x*(-4-1x*(-4-x)*2+5)+1+1*5)=0

We add all the numbers together, and all the variables

-1x*(-4-14x*(-4-1x*(-1x-4)*2+5)+1+1*5)=0

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