Let , smallest integer is x.
So , other one is x + 1.
By , given conditions :
[tex]x( x + 1 ) = 8x + 18\\\\x^2+x=8x+18\\\\x^2-7x-18=0\\\\x^2-9x+2x-18=0\\\\x(x-9)+2(x-9)=0\\\\x=-2 \ ,\ x = 9[/tex]
Since, the numbers are positive so, x = -2 is ignored.
Therefore, the numbers are 9 and 10.
Hence, this is the required solution.