Question 639550
To solve, we must set this question up as equations.  The first part of the  problem says that the sum of 3 numbers is 79.  In other words

x + y + z = 79

The second part of the problem says that the second number is 9 times the first.  In other words

y = 9x, where y is the second number, and x is the first number from the first equation we set up

The third part of the problem says that the third number is 3 more than the second number.  In other words

z = y + 3

To solve, we have to find each number one at a time.  So, in the first equation we set up   x + y + z = 79, the easiest number to find first is x.  So, we will substitute y and z in the first equation with y and z in the second and third equations.  Since y = 9x, in our first equation, we will substitute y with 9x, and since z = y + 3, and we know that y = 9x, we will substitute z in our first equation with 9x + 3.  So, our first equation now looks like

x + 9x + 9x + 3 = 79

Combining like terms on the left side of the equal sign gives us

19x + 3 = 79

Now, subtract 3 from both sides:

19x + 3 - 3 = 79 - 3

Which equals

19x = 76

Divide 19 by both sides so that we get x all by itself:

19x/19 = 76/19

Which gives us our first number:

x = 4

Now, we know that y = 9x, so replace x with 4: y = 9 x 4 = 36.  So, our second number is 36.

We know that z is 3 more than y, so replace y with 36 in our 3rd equation:  z = 36 + 3 = 39.  Our third number, therefore is 39

Final Answer:  Our three numbers are 4, 36, and 39