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Find three consecutive even integers. Such that 7 times of first integers is 14 more than the sum of second and third integers.

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TylersWife420

Answer:

no clue sorry I couldn't be of more help

vidalmari6p99qfk

Answer:

4, 6 and 8

Step-by-step explanation:

1st number       x

2nd number   x+2

3rd number     x+4

7x = (x+2) + (x+4) +14

7x = 2x+20

5x=20

x=4

7*4=28          (6+8)+14 =14+14 = 28

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