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(x+2)(2x-8+x+2)=(x+1)(2x-5+x+1)
We move all terms to the left:
(x+2)(2x-8+x+2)-((x+1)(2x-5+x+1))=0
We add all the numbers together, and all the variables
(x+2)(3x-6)-((x+1)(3x-4))=0
We multiply parentheses ..
(+3x^2-6x+6x-12)-((x+1)(3x-4))=0
We calculate terms in parentheses: -((x+1)(3x-4)), so:We get rid of parentheses
(x+1)(3x-4)
We multiply parentheses ..
(+3x^2-4x+3x-4)
We get rid of parentheses
3x^2-4x+3x-4
We add all the numbers together, and all the variables
3x^2-1x-4
Back to the equation:
-(3x^2-1x-4)
3x^2-3x^2-6x+6x+1x-12+4=0
We add all the numbers together, and all the variables
x-8=0
We move all terms containing x to the left, all other terms to the right
x=8
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