Three cousins have ages that are consecutive integers. the product of the two older counsins ages is twelve less than six times the sum of the younger two cousins ages. write an equation and solve to find the three cousins ages algebraically.

N, n+1, n+2

(n+1)(n+2)=6(n+n+1)-12

n^2+3n+2=12n+6-12

n^2+3n+2=12n-6

n^2-9n+8=0

n^2-n-8n+8=0

n(n-1)-8(n-1)=0

(n-8)(n-1)=0

So there are actually two solutions that satisfy the conditions for the cousin's ages...

1,2, and 3  years old and 8,9,and 10 years old.

check...

2*3=6(1+2)-12,  6=18-12, 6=6

9*10=6(8+9)-12,  90=102-12, 12=12