Question

Find 3 consecutive integers such that the product of the two smaller integers is 38 more than twice the largest. :)

Answer 1

If the integers are consecutive, then you only need one variable to describe each of them: \(x, x+1,x+2\).
The product of the two smaller ones is 38 more than twice the largest. The product of the two smaller ones is \[(x)(x+1)\] and 38 more than twice the largest is \[2(x+2)+38\]

Answer 2

z-y = 1
y-x = 1
xy = 2z + 38

y= x+ 1
z = 2 + x
sub these into third equation
x(x+1) = 2(2+x) + 38
x^2 + x = 4 + 2x + 38
x^2 - x -42 = 0
x = 6

so three numbers are 6, 7 and 8

Answer 3

yep, I completely misread the problem >.<