-3x+4x=4+5(2x+1

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Solution for -3x+4x=4+5(2x+1 equation: 



-3x+4x=4+5(2x+1

We move all terms to the left:

-3x+4x-(4+5(2x+1)=0

We add all the numbers together, and all the variables

x-(4+5(2x+1)=0



We calculate terms in parentheses: -(4+5(2x+1), so:

4+5(2x+1

determiningTheFunctionDomain
5(2x+1+4

Back to the equation:

-(5(2x+1+4)



We add all the numbers together, and all the variables

x-(5(2x+5)=0



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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





We calculate terms in parentheses: -(5(2x+5), so:

5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





We calculate terms in parentheses: -(5(2x+5), so:

5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





We calculate terms in parentheses: -(5(2x+5), so:

5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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5(2x+5





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