Find four consecutive multiples of three that the product of the first and third miners minus eight times the fourth number is negative 48

The four consecutive multiples of three are 6, 9, 12, and 15

Step-by-step explanation:

let the first multiple of 3 = n

second multiple of 3 = (n + 3)

third multiple of 3 =(n + 6)

fourth multiple of 3 =(n + 9)

product of the first and the third = n(n + 6) = n² + 6n

eight times the fourth number = 8(n + 9) = 8n  + 72

The difference between the two =  n² + 6n - (8n  + 72) = -48

n² + 6n - 8n  - 72 = -48

n² - 2n =  -48 + 72

n² - 2n = 24

n² - 2n - 24 = 0

n²  - 2n - 24 = 0

n² + 4n - 6n - 24 = 0

n(n + 4) - 6(n + 4) = 0

(n - 6)(n + 4) = 0

n = 6    or  - 4

n should be a positive number, thus n = 6

The second multiple of 3 =  (n + 3) = 6 + 3 = 9

The third multiple of 3 =(n + 6) = 6 + 6 = 12

The fourth multiple of 3 = (n + 9) = 6 + 9 = 15

yes yes yes yes yes yes yes yes yes

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: