In this problem the letters x and y are each one digit numbers. Compare x+y O 19?

Do I put any number for x and y ? please help.

Your answer was no but how do I do this problem?

I'm sorry, I mean <,>,or =. Thank You for your help.

Can any 1-digit numbers equal 19?

The largest 1-digit numbers are 8 and 9.

Can any 2 one-digit numbers when added equal 19?

So you can use any 1 digit number. For example: I used 2+3. Since 2+3=5 what I did was then re-write my problem to 5 < 9. In the circles, I used a less than sign since 5 is less than 9. But you can use any 1 digit number to add.

To solve this problem, we need to compare the sum of two one-digit numbers, x and y, with 19.

To find the answer, you need to consider all possible combinations of x and y that are one-digit numbers. Since x and y can be any one-digit number, you can systematically evaluate each possible combination.

Here's how you can do it:

1. Start with the smallest possible values for x and y. In this case, both x and y can be 0 since 0 is the smallest one-digit number.
- Evaluate x + y: 0 + 0 = 0. Since 0 is not greater than 19, the statement x + y > 19 is false.
- In this case, x + y is not greater than 19, so we can conclude that x + y is not greater than 19.

2. Now, move on to the next possible combination of values.
- x = 0, y = 1
- Evaluate x + y: 0 + 1 = 1. Since 1 is not greater than 19, the statement x + y > 19 is false.

3. Repeat this process for all possible combinations of x and y, incrementing their values.
- For example, next you would consider x = 0, y = 2, then x = 0, y = 3, and so on.
- Keep evaluating x + y and checking if it is greater than 19. If it is, you can conclude that x + y is greater than 19. If not, continue with the next combination.

Continue this process until you have evaluated all possible combinations of x and y. By comparing x + y with 19 for each combination, you can determine if x + y is greater than, less than, or equal to 19.