1/5(10+-5(x+-2))=1/10(x+1)

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Solution for 1/5(10+-5(x+-2))=1/10(x+1) equation:



1/5(10+-5(x+-2))=1/10(x+1)
We move all terms to the left:
1/5(10+-5(x+-2))-(1/10(x+1))=0
Domain of the equation: 5(10+-5(x+-2))!=0
x∈R
Domain of the equation: 10(x+1))!=0
x∈R
We add all the numbers together, and all the variables
1/5(10+-5(x-2))-(1/10(x+1))=0
We use the square of the difference formula
1/5(10-5(x-2))-(1/10(x+1))=0
We calculate fractions
(10xx/(5(10-5(x-2))*10(x+1)))+(-5x1/(5(10-5(x-2))*10(x+1)))=0
We calculate terms in parentheses: +(10xx/(5(10-5(x-2))*10(x+1))), so:
10xx/(5(10-5(x-2))*10(x+1))
We multiply all the terms by the denominator
10xx
Back to the equation:
+(10xx)
We calculate terms in parentheses: +(-5x1/(5(10-5(x-2))*10(x+1))), so:
-5x1/(5(10-5(x-2))*10(x+1))
We multiply all the terms by the denominator
-5x1
We add all the numbers together, and all the variables
-5x
Back to the equation:
+(-5x)
We get rid of parentheses
10xx-5x=0
We add all the numbers together, and all the variables
-5x+10xx=0

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