Posted by **Please help** on Monday, May 31, 2010 at 9:49pm.

Here is the problem:

A two-digit number is eight times the sum of its digits.When the number is added to the number obtained by reversing the digits,the sum is 99. Find the original number.

Please explain hot to solve it and it would help a lot to if you explain other algebra word problems like:

Distance rate and time problems,Mixture Problems,wind and water current problems,work problem,area problem,cost and value problems if you could explain or give a link to something helpful,

Thanks

- Algebra -
**Damon**, Monday, May 31, 2010 at 9:56pm
nm

write two digit number nm as m + 10 n

m + 10n = 8 (n+m)

and

m + 10n + n + 10 m = 99

then solve the two equations

2 n = 7 m

11n+11m = 99 or n+m=9

n = (7/2)m

(7/2)m + m = 9

7 m + 2 m =18

9 m = 18

m = 2

n = 7

number = 72

- Algebra -
**MathMate**, Monday, May 31, 2010 at 9:59pm
This is a question with two equations, x, y each representing one of the two digits.

the number is equal to 8 times the sum of its digits, therefore:

10x+y = 8(x+y), or

2x=7y

x=3.5y

That leaves only one solution: x=7, y=2.

When the number is added to that with the digits reversed, it should add up to 99:

72+27=99 OK.

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