1=1/2(x+3)-5

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Solution for 1=1/2(x+3)-5 equation:



1=1/2(x+3)-5
We move all terms to the left:
1-(1/2(x+3)-5)=0
Domain of the equation: 2(x+3)-5)!=0
x∈R
We multiply all the terms by the denominator
-(1+1*2(x+3)-5)=0
We calculate terms in parentheses: -(1+1*2(x+3)-5), so:
1+1*2(x+3)-5
determiningTheFunctionDomain 1*2(x+3)+1-5
We add all the numbers together, and all the variables
1*2(x+3)-4
Wy multiply elements
2x(x-4
Back to the equation:
-(2x(x-4)
We calculate terms in parentheses: -(2x(x-4), so:
2x(x-4
We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
2x(x-4
We calculate terms in parentheses: -(2x(x-4), so:
2x(x-4
We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
2x(x-4
We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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We calculate terms in parentheses: -(2x(x-4), so:
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