One number is twice as large as another and the third number is 11 more then the smaller one if the sum of the three numbers is 79 find the smallest number?

Let the numbers be x, y and z,
Where y is the smallest number, so,

By question,
1st case,
x=2y eq(1)

2nd case,
z=y+11 eq(2)

Finally, 3rd case,
x+y+z=79 eq(3)

Equating the values of x and z from eq(1) and eq(2) in eq(3), we get,

(2y)+y+(y+11)=79
Or, 2y+y+y+11=79
Or, 4y=79-11
Or, 4y=68
Or, y=68/4
.•. y=17,,

Hence, the smallest number is 17.

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(-0, -7) is your answer