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
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
