Question

question 10 points 1
the difference between two numbers is greater than the first number and the sum of the two numbers is greater than 21. find the system of inequalities representing the given situation.

\\(\bigcirc\\) \\(x + y < 21, x - y < x, x \geq 0\\) and \\(y \geq 0\\)

\\(\bigcirc\\) \\(x + y > 21, x - y > x, x \geq 0\\) and \\(y \geq 0\\)

\\(\bigcirc\\) \\(x + y > 21, x - y > x, x > 0\\) and \\(y > 0\\)

\\(\bigcirc\\) \\(x + y > 21, x - y > x, x < 0\\) and \\(y < 0\\)

Explanation:

Step1: Analyze the sum condition

The problem states that the sum of the two numbers is greater than 21. If we let the two numbers be \(x\) and \(y\), then the sum \(x + y>21\).

Step2: Analyze the difference condition

The difference between two numbers (let's say \(x-y\)) is greater than the first number \(x\). So \(x - y>x\). If we subtract \(x\) from both sides of the inequality \(x - y>x\), we get \(-y>0\), which implies \(y < 0\)? Wait, no, wait. Wait, let's re - examine. Wait, if \(x-y>x\), subtract \(x\) from both sides: \(x - y-x>x - x\), which simplifies to \(-y>0\), or \(y < 0\). But also, we need to consider the nature of numbers. But let's check the non - negativity. Wait, but in the context of numbers (usually, when we talk about numbers in such problems, if not specified, they can be non - negative, but let's check the options.

Wait, the first part: sum is greater than 21, so \(x + y>21\) (so we can eliminate the first option which has \(x + y<21\)).

Now, the difference: \(x - y>x\). Let's solve \(x - y>x\). Subtract \(x\) from both sides: \(-y>0\), so \(y < 0\)? Wait, but let's check the options. Wait, maybe I made a mistake. Wait, maybe the two numbers are positive? No, wait, let's check the options.

Wait, the second option has \(x\geq0\) and \(y\geq0\), the third has \(x > 0\) and \(y>0\), the fourth has \(x < 0\) and \(y < 0\).

Wait, let's re - do the difference condition. The difference between two numbers is greater than the first number. Let the two numbers be \(x\) and \(y\). So \(x - y>x\) (assuming the first number is \(x\) and the second is \(y\), and we are taking \(x - y\) as the difference). Then \(x - y>x\) implies \(-y>0\) or \(y < 0\). But also, the sum \(x + y>21\). If \(y < 0\) and \(x + y>21\), then \(x>21 - y\). Since \(y < 0\), \(-y>0\), so \(x>21+|y|\). But let's check the options.

Wait, the fourth option has \(x < 0\) and \(y < 0\). Let's check the sum: if \(x<0\) and \(y < 0\), then \(x + y<0\), which cannot be greater than 21. So the fourth option is out.

The third option has \(x>0\) and \(y>0\), but if \(x>0\) and \(y>0\), and \(x - y>x\), then \(-y>0\) which would mean \(y < 0\), a contradiction.

The second option has \(x\geq0\) and \(y\geq0\), but again, \(x - y>x\) would imply \(y < 0\), a contradiction? Wait, no, maybe I assigned the numbers wrong. Maybe the difference is \(y - x>x\)? Wait, the problem says "the difference between two numbers is greater than the first number". Let's assume the two numbers are \(x\) (first) and \(y\) (second). So difference is \(x - y\) (if \(x>y\)) or \(y - x\) (if \(y>x\)). But the problem says "the difference between two numbers is greater than the first number". So if the first number is \(x\), then \(|x - y|>x\)? No, the problem says "the difference between two numbers is greater than the first number", so \(x - y>x\) (assuming the difference is \(x - y\)).

Wait, let's check the options again. The second option: \(x + y>21\), \(x - y>x\), \(x\geq0\) and \(y\geq0\). But \(x - y>x\) implies \(y < 0\), which contradicts \(y\geq0\). The third option: \(x + y>21\), \(x - y>x\), \(x>0\) and \(y>0\). Again, \(x - y>x\) implies \(y < 0\), contradiction. The fourth option: \(x + y>21\), \(x - y>x\), \(x < 0\) and \(y < 0\). But if \(x<0\) and \(y < 0\), \(x + y<0\), which is not greater than 21. Wait, this is confusing. Wait, maybe I made a mistake in the difference. Maybe the difference is \(y - x>x\) (if the first number is \(x\) and the second is \(y\), and the difference is \(y - x\)). Let's try that. If \(y - x>x\), then \(y>2x\). And the sum \(x + y>21\). But the options don't have this. Wait, the original options:

Option 2: \(x + y>21\), \(x - y>x\), \(x\geq0\) and \(y\geq0\)

Option 3: \(x + y>21\), \(x - y>x\), \(x>0\) and \(y>0\)

Option 4: \(x + y>21\), \(x - y>x\), \(x < 0\) and \(y < 0\)

Wait, maybe the problem has a typo, or maybe I misinterpret the difference. Wait, maybe the two numbers are \(x\) and \(y\), and the difference is \(y - x\) (the second minus the first) is greater than the first number \(x\). So \(y - x>x\), which is \(y>2x\). But the options don't have this. Wait, no, the options have \(x - y>x\). Let's go back.

If \(x - y>x\), then \(x - y-x>0\), \(-y>0\), \(y < 0\). And \(x + y>21\). If \(y < 0\), then \(x>21 - y\). Since \(y < 0\), \(-y>0\), so \(x>21+|y|\). Now, for \(x + y>21\) and \(y < 0\), \(x\) must be positive (because \(y\) is negative, and \(x>21 - y\), and \(-y>0\), so \(x>21\)). But in the options, the fourth option has \(x < 0\) and \(y < 0\), which would make \(x + y<0\), not greater than 21. The third option has \(x>0\) and \(y>0\), but \(y < 0\) from \(x - y>x\), contradiction. The second option has \(x\geq0\) and \(y\geq0\), contradiction. Wait, this is a problem. Wait, maybe the difference is \(y - x>x\) (the other way around). Let's try \(y - x>x\), so \(y>2x\). And \(x + y>21\). If \(x\geq0\) and \(y\geq0\), then this is possible. But the options have \(x - y>x\). Wait, maybe the problem meant the difference between the two numbers (absolute value) but no. Wait, let's check the options again.

Wait, the first option is out because \(x + y<21\) is wrong. The fourth option: \(x < 0\) and \(y < 0\), sum is negative, wrong. The third option: \(x>0\), \(y>0\), but \(x - y>x\) implies \(y < 0\), wrong. The second option: \(x\geq0\), \(y\geq0\), but \(x - y>x\) implies \(y < 0\), wrong. Wait, this must be a mistake. Wait, no, maybe I made a mistake in solving \(x - y>x\). Let's do it again:

\(x - y>x\)

Subtract \(x\) from both sides: \(x - y-x>x - x\)

\(-y>0\)

Multiply both sides by - 1 (and reverse the inequality): \(y < 0\)

So \(y\) is negative. Now, the sum \(x + y>21\). Since \(y\) is negative, \(x>21 - y\). But \(y\) is negative, so \(-y\) is positive, so \(x>21+|y|\). So \(x\) must be positive (because \(21+|y|>21>0\)). So \(x>0\) and \(y < 0\). But none of the options have \(x>0\) and \(y < 0\). Wait, the options are:

1. \(x + y<21,x - y<x,x\geq0\) and \(y\geq0\)

2. \(x + y>21,x - y>x,x\geq0\) and \(y\geq0\)

3. \(x + y>21,x - y>x,x>0\) and \(y>0\)

4. \(x + y>21,x - y>x,x < 0\) and \(y < 0\)

Wait, maybe the problem has a mistake, but among the given options, the only one with \(x + y>21\) and \(x - y>x\) and considering the fact that if we ignore the contradiction in non - negativity (maybe the problem has a typo and the difference is \(y - x>x\)) but among the options, the second option is the closest in terms of having \(x + y>21\) and \(x - y>x\) with non - negative numbers (even though there is a contradiction in \(y\)'s sign). Wait, no, maybe the original problem had a different setup. Wait, maybe the two numbers are \(x\) and \(y\), and the first number is \(y\) and the second is \(x\). Let's try: difference between two numbers (\(x - y\)) is greater than the first number (\(y\)). So \(x - y>y\), which is \(x>2y\). And sum \(x + y>21\). But the options don't have this.

Wait, I think there is a mistake in my analysis. Let's check the options again. The second option has \(x + y>21\), \(x - y>x\), \(x\geq0\) and \(y\geq0\). Even though \(x - y>x\) implies \(y < 0\), maybe the problem assumes that the numbers can be such that we have these inequalities regardless of the sign contradiction, or maybe I misread the difference. Maybe the difference is \(y - x>x\) (the other number minus the first is greater than the first). Let's try \(y - x>x\), so \(y>2x\). And \(x + y>21\). If \(x\geq0\) and \(y\geq0\), this is possible. But the options have \(x - y>x\).

Wait, maybe the problem is correct, and the answer is the second option, even with the contradiction, because it's the only one with \(x + y>21\) and \(x - y>x\) and non - negative (or semi - non - negative) numbers.

Answer:

B. \(x + y>21,x - y>x,x\geq0\) and \(y\geq0\) (Here, we assume that despite the contradiction in the sign of \(y\) from \(x - y>x\), this is the intended option as it has the correct sum inequality and the correct difference inequality structure among the given options.)