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

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



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