Let's assign some variable:
x = smaller integer
y = bigger integer
We can set this up as an expression:
[tex]y = 8x[/tex]
[tex]xy = 128[/tex]
Solve for a variable in the first equation, y is already given, so plug that into the second equation:
[tex](8x)(x) = 128[/tex]
[tex]8x^2=128[/tex]
[tex]x^2 = 16[/tex]
[tex]x = 4, -4[/tex]
Since we only care about the positive values (scalar multiplication), we can rule out the -4, and we are left with x = 4. y is 8 times x, so it has to be 32. Your final answer: x = 4, y = 32
