Question

Eleven times one number is 9 more than 8 times another number. The first number minus four is half the second number. Enter the smaller of the two numbers.

Answer 1

Let the numbers are: x and y

Eleven times one number is 9 more than 8 times another number

so,

`11x=8y+9\rightarrow(1)`

And, The first number minus four is half the second number

so,

`x-4=0.5y\rightarrow(2)`

from equation (2):

`x=0.5y+4`

substitute with (x) into equation (1) then solve for (y):

`\begin{gathered} 11(0.5y+4)=8y+9 \\ 5.5y+44=8y+9 \\ 5.5y-8y=9-44 \\ -2.5y=-35 \\ \\ y=(-35)/(-2.5)=14 \end{gathered}`

substitute with (y) into equation (2) to find (x):

`\begin{gathered} x-4=0.5\cdot14 \\ x-4=7 \\ x=7+4 \\ x=11 \end{gathered}`

so, the answer will be:

The smaller of the two numbers = 11