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prohojiy [21]
4 years ago
13

Math problem is on the attached picture. Thanks!

Mathematics
1 answer:
34kurt4 years ago
8 0

Answer:

the consecutive integars is 11,12,13,14,15

Step-by-step explanation:

  • let the first integar  be described as y
  • since they are consecutive ( following each other)
  • the second will y + 1
  • the third is y + 2
  • the fourth is y + 3
  • the fifth y + 4

then it says the sum of the first and 4 times the third = 60 less than 3 times the sum of the second, fourth and fifth

60 less than 3 times the sum of the second, fourth and fifth means that 3 times the sum of the second, fourth and fifth minus 60

y+ 4(y+2) = 3{ (y +1) +(y+3) + (y+4)} -60

  • open the bracket

y + 4y + 8 =3[ y +1 + y + 3 +y+4} - 60

5y +8=3(y+y+y+1+3+4)-60

5y+8=3(3y+8)-60

  • simplify and open the brackets

5y+8=9y+24-60

5y+8=9y-36

  • subtract 8 from both sides

5y+8-8=9y-36-8

5y=9y-44

  • subtact 9y from both sides

5y-9y=9y-9y-44

-4y=-44

  • divide -4 from both sides

-4y/-4=-44/-4 ( minus divided by minus is plus

y = 11

  • the first integar is y = 11
  • the second will y + 1 = 11+1 = 12
  • the third is y + 2 = 11 +2 = 13
  • the fourth is y + 3 = 11 + 3 = 14
  • the fifth y + 4 = 11 + 4 = 15
  • the consecutive integars is 11,12,13,14,15
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