Find 3 consecutive integers such that the product of -5 and the sum of the first two is 43 less than twice the

Question

second. (Please show your work. Thank you!)

Answer 1

write some equations :)

let x = 1st #
x+1 = 2nd consecutive #
x+2 = 3rd consecutive #

-5 (x + x+1) = 2(x+1) - 43

-5x -5x -5 = 2x + 2 - 43

-10x - 2x = -36

-12 x = -36

x = 3

x+1 = 4

x+2 = 5

check: -5(3+4) = 2(4) - 43

-35 = -35 check!! :)

Answer 2

(-5) (x + x+1) + 43 = 2 (x + 1)
(-5) (2X + 1) + 43 = 2x + 2
-10x - 5 + 43 = 2x + 2
-10x + 38 = 2x + 2
36 = 12x
x = 3

so the integers are 3, 4, and 5

to double check, (-5) * (3 + 4) = 2(4) - 43
(-5)*(7) = 8 - 43
-35 = - 35