3(2k+11)=1/4(16k+24)+2k

Simple and best practice solution for 3(2k+11)=1/4(16k+24)+2k equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

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Solution for 3(2k+11)=1/4(16k+24)+2k equation: 



3(2k+11)=1/4(16k+24)+2k

We move all terms to the left:

3(2k+11)-(1/4(16k+24)+2k)=0



Domain of the equation: 4(16k+24)+2k)!=0

k∈R



We multiply parentheses

6k-(1/4(16k+24)+2k)+33=0

We multiply all the terms by the denominator


6k*4(16k+24)+2k)-(1+33*4(16k+24)+2k)=0

Wy multiply elements

24k^2(1+2k)-(1+132k(1+2k)=0



We calculate terms in parentheses: -(1+132k(1+2k), so:

1+132k(1+2k

determiningTheFunctionDomain
132k(1+2k+1

Back to the equation:

-(132k(1+2k+1)



We add all the numbers together, and all the variables

24k^2(2k+1)-(132k(2k+2)=0



We calculate terms in parentheses: -(132k(2k+2), so:

132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2





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132k(2k+2

				

			
	
			

			
			

							

									
									
									
									
									
									

														
			

			
				

						
				
				
					
			
						
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