Answered

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2. Find four consecutive integers such that six times the sum of the first and third is 12
greater than 9 times the fourth.

Sagot :

Answer:

9

Step-by-step explanation:

a,a+1,a+2,a+3,...

6(a+(a+2))=12+9(a+3)

6(2a+2)=12+9a+27

12a+12=12+9a+27

12a-9a=27

3a=27

a=9

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