The sum of the three digits is 9. The tens digit is 1 more than the hundreds digit...

[asdfasdfasdf]

The sum of the three digits is 9. The tens digit is 1 more than the hundreds digit. When the digits are reversed, the new number is 99 less than the original number. Find the original number.

 

[asdfasdfasdf]

Some work i did.

I did the number as abc
so a+b+c=9
and a+(a+1)+c=9
so 2a+c=8
also I have cba+99=abc
That's all I have.

 

[asdfasdfasdf]

I don't think that is the answer to my question... Thanks Anyways.

 

[Harry_the_cat]

I did the number as abc
so a+b+c=9
and a+(a+1)+c=9
so 2a+c=8
also I have cba+99=abc
That's all I have.

abc means a*b*c. You should use 100a +10b +c if a is the first digit (100s), b is the second digit (10s) and c is the third digit (units).

When the digits are reversed you'll have 100c + 10b +a.

See if you can go from there.

 

[asdfasdfasdf]

Finished!

Thanks Harry the Cat, helped a lot.
My final work :
Let the number be abc
a+b+c=9
a+(a+1)+c=9
2a+c=8
100a+10b+c-99=100c+10b+a
99a-99=99c
Substitute (bolded)
99a-99=-198a+792
-891=-297a
a=3
Going back to top (underlined)
3+3+1+c=9
7+c=9
c=2
b=4
So the number is 342

 

[Harry_the_cat]

Thanks Harry the Cat, helped a lot.
My final work :
Let the number be abc ... this statement is still incorrect
a+b+c=9
a+(a+1)+c=9
2a+c=8
100a+10b+c-99=100c+10b+a
99a-99=99c
Substitute (bolded)
99a-99=-198a+792
-891=-297a
a=3
Going back to top (underlined)
3+3+1+c=9
7+c=9
c=2
b=4
So the number is 342

Great! Did you check that your answer of 342 satisfies all the conditions in the original question? This way you'll know you are correct.

 

[stapel]

The sum of the three digits is 9. The tens digit is 1 more than the hundreds digit. When the digits are reversed, the new number is 99 less than the original number. Find the original number.

Go "see how" here:
https://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq.question.995473.html

I don't think that is the answer to my question... Thanks Anyways.

Then maybe try this lesson, which reflects the methodology you were given here.

 

[XDUIGOKU]

Finished!

Thanks Harry the Cat, helped a lot.
My final work :
Let the number be abc
a+b+c=9
a+(a+1)+c=9
2a+c=8
100a+10b+c-99=100c+10b+a
99a-99=99c
Substitute (bolded)
99a-99=-198a+792
-891=-297a
a=3
Going back to top (underlined)
3+3+1+c=9
7+c=9
c=2
b=4
So the number is 342

This is the right answer because the tens digit is 2x the hundredths digit and the units digit is 1 more than the hundredths digit and they add up to 9

 

[lookagain]

This is the right answer because the tens digit is 2x the hundredths digit and the units digit is 1 more than the hundredths digit and they add up to 9

The problem mentions the "hundreds digit," not the "hundredths digit." And, no,
the tens digit is not two times the hundreds digit. ** Also, the units digit is not one
more than the hundreds digit. The three digits do add up to nine, but you did not
address checking the new number being 99 less than the original number.


** This is not one of the problem's conditions.