3/4(2x+5)=5(x+12)

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Solution for 3/4(2x+5)=5(x+12) equation: 



3/4(2x+5)=5(x+12)

We move all terms to the left:

3/4(2x+5)-(5(x+12))=0



Domain of the equation: 4(2x+5)!=0

x∈R



We multiply all the terms by the denominator


-((5(x+12)))*4(2x+5)+3=0



We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





We calculate terms in parentheses: -((5(x+12)))*4(2x+5), so:

(5(x+12)))*4(2x+5





							

									
									
									
									
									
									

														
			

			
				

						
				
				
					
			
						
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