SOLUTION: the sum of 3 numbers is 79 the second number is 9 times the first nd the third number is three more then the second find the numbers

Question 639550: the sum of 3 numbers is 79 the second number is 9 times the first nd the third number is three more then the second find the numbers
Answer by josh_jordan(263) (Show Source): You can put this solution on YOUR website!
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

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