Three consecutive integers are such that 8 less than the first number plus the second times 6 plus half the third is -31. Find the numbers

Answer 1

let first number be : a

let 2nd number be : b

let 3rd number be : c,

Three consecutive integers ---->a, b= a+1, c=a+2

8 less than the first number ----> a-8

plus ----> +

the second times 6 ----> 6*b

plus----> +

half the third ----> (1/2)*c

is ----> =

-31

a-8 + 6*b+ (1/2)*c =-31

a-8+ ( 6*(a+1))+ (1/2)*(a+2)=-31

a-8 + 6a +6 + (a/2 ) +1 =-31

put a's together: 7.5a -1=-31----> a=-30/7.5= -4

so a= -4

b=-3

c=-2