1(9-x)+x=10x+(9-x)+27

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Solution for 1(9-x)+x=10x+(9-x)+27 equation: 



1(9-x)+x=10x+(9-x)+27

We move all terms to the left:

1(9-x)+x-(10x+(9-x)+27)=0

We add all the numbers together, and all the variables

1(-1x+9)+x-(10x+(-1x+9)+27)=0

We add all the numbers together, and all the variables

x+1(-1x+9)-(10x+(-1x+9)+27)=0



We calculate terms in parentheses: -(10x+(-1x+9)+27), so:

10x+(-1x+9)+27

We get rid of parentheses

10x-1x+9+27

We add all the numbers together, and all the variables

9x+36

Back to the equation:

-(9x+36)



We get rid of parentheses

x+1(-1x+9)-9x-36=0

We add all the numbers together, and all the variables

-8x+1(-1x+9)-36=0

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

-8x+1(-1x+9)=36

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