Let the numbers are: x and y
Eleven times one number is 9 more than 8 times another number
so,
[tex]11x=8y+9\rightarrow(1)[/tex]And, The first number minus four is half the second number
so,
[tex]x-4=0.5y\rightarrow(2)[/tex]from equation (2):
[tex]x=0.5y+4[/tex]substitute with (x) into equation (1) then solve for (y):
[tex]\begin{gathered} 11(0.5y+4)=8y+9 \\ 5.5y+44=8y+9 \\ 5.5y-8y=9-44 \\ -2.5y=-35 \\ \\ y=\frac{-35}{-2.5}=14 \end{gathered}[/tex]substitute with (y) into equation (2) to find (x):
[tex]\begin{gathered} x-4=0.5\cdot14 \\ x-4=7 \\ x=7+4 \\ x=11 \end{gathered}[/tex]so, the answer will be:
The smaller of the two numbers = 11
