Digits 6

[harpazo]

The sum of the digits of a 2 digit number is 9. If the digits are reversed, the new number is 9 less than 3 times the original number. Find the original number.

Solution:

Let r = tens digit
Let s = ones digit

r + s = 9
10s + r = 3(r + s) - 9

Is this the correct set up?

Next week, I will post distance word problems as I make my way through algebra applications followed by geometry and then trigonometry applications. Lots of practice with word problems just ahead. My main concern, as I've stated many times, is the set up.

 

[Deleted member 4993]

The sum of the digits of a 2 digit number is 9. If the digits are reversed, the new number is 9 less than 3 times the original number. Find the original number.

Solution:

Let r = tens digit
Let s = ones digit

r + s = 9
10s + r = 3(r + s) - 9

Is this the correct set up?

Next week, I will post distance word problems as I make my way through algebra applications followed by geometry and then trigonometry applications. Lots of practice with word problems just ahead. My main concern, as I've stated many times, is the set up.

Second equation is incorrect.

Let the original number N_o = 10*r + s

Let the "reversed" number N_R = 10*s + r

Then we have:

If the digits are reversed, the new number is 9 less than 3 times the original number.

So we have:

N_R = 3 * N_o - 9 \(\displaystyle \ \ \to \ \ \ \) 10*s + r = 3 * (10*r + s) - 9

29*r - 7*s = 9 ................................. This should be your second equation

 

[harpazo]

Second equation is incorrect.

Let the original number N_o = 10*r + s

Let the "reversed" number N_R = 10*s + r

Then we have:

If the digits are reversed, the new number is 9 less than 3 times the original number.

So we have:

N_R = 3 * N_o - 9 \(\displaystyle \ \ \to \ \ \ \) 10*s + r = 3 * (10*r + s) - 9

29*r - 7*s = 9 ................................. This should be your second equation

I'll take it from here.

 

[Deleted member 4993]

The sum of the digits of a 2 digit number is 9. If the digits are reversed, the new number is 9 less than 3 times the original number. Find the original number.

To complete the solution (for future reference):

Let the original number N_o = 10*r + s

Let the "reversed" number N_R = 10*s + r

Then we have:

If the digits are reversed, the new number is 9 less than 3 times the original number.

So we have:

N_R = 3 * N_o - 9 → → 10*s + r = 3 * (10*r + s) - 9

29*r - 7*s = 9 ....................................................................................(1) and

r + s = 9 ....\(\displaystyle \to \ \ \ \ \) 7 * r + 7 * s = 63 .................................(2)

Add (1) and (2) to get

36 * r = 72 ...\(\displaystyle \to \ \ \ \ \) r = 2 ,,,,,,,,,,...\(\displaystyle \to \ \ \ \ \) s = 7

Original number = 10*2 + 7 = 27