help algebra

 

 

Find three consecutive odd integers such that the sum of the first, two times the second and three times the third is 34.     

 

Call the 1^st integer (x)  

Call the 2^nd integer (x + 2)  

Call the 3^rd integer (x + 4)  

 

Given                                                  (x) + 2(x + 2) + 3(x + 4)  =  34  

Now, just multiply it out                         x + 2x + 4 + 3x + 12  =  34   

Combine like terms                              6x + 16  =  34  

Subtract 16 from both sides                6x  =  18  

Divide both sides by 6                          x  =  3  

 

The 1^st integer is 3, the 2^nd integer is 5, and the 3^rd integer is 7  

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