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1/5(2x+13)=4(4-4x)+3
We move all terms to the left:
1/5(2x+13)-(4(4-4x)+3)=0
Domain of the equation: 5(2x+13)!=0We add all the numbers together, and all the variables
x∈R
1/5(2x+13)-(4(-4x+4)+3)=0
We multiply all the terms by the denominator
-((4(-4x+4)+3))*5(2x+13)+1=0
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
We calculate terms in parentheses: -((4(-4x+4)+3))*5(2x+13), so:
(4(-4x+4)+3))*5(2x+13
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We cal
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