One number is 9 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 639​, find the numbers.

Answer 1

n = first number

9n = second number

100 + n = third number

n + 9n + (100 + n) = 639

11n = 539

n = 49

You should be able to finish from here. Test your final answers.

Answer 2

To solve we will start with the given parameters. I will use the variables x (first number), y (second number), and z (third number).

Based on the above we know the following

y=9x (One number is 9 times a first number).

z=100+x (A third number is 100 more than the first number)

x+y+z=639 (the sum of the three numbers is 639​)

we will then substitute the values of y & z in the last equation

x+y+z=639

x + 9x + (100+x) = 639

10x+100+x = 639

11x+100=639

11x+100-100=639-100

11x=539

11x/11=539/11

x=49

y=9x

y=9(49)=441

z=100+x = 100+49 = 149