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If x and y are integers in the above inequalities
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27 Feb 2026, 14:45
27 Feb 2026, 14:45
Expert Reply
00:00
Question Stats:
54% (02:56) correct
45% (02:28) wrong based on 11 sessions
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$$
\(\begin{gathered}
x^2-y^2<8 \\
x+y>3
\end{gathered}\)
$$
If \(x\) and \(y\) are integers in the above inequalities and \(0<y<x\), what is the greatest possible value of \(x\) ?
A. 1
B. 2
C. 3
D. 4
E. 5
\(\begin{gathered}
x^2-y^2<8 \\
x+y>3
\end{gathered}\)
$$
If \(x\) and \(y\) are integers in the above inequalities and \(0<y<x\), what is the greatest possible value of \(x\) ?
A. 1
B. 2
C. 3
D. 4
E. 5
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If x and y are integers in the above inequalities
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02 Mar 2026, 02:41
02 Mar 2026, 02:41
Expert Reply
Step-by-Step Solution:
1. Factor the first inequality:
Recall the difference of squares: $\(x^2-y^2=(x-y)(x+y)\)$.
So, the inequality becomes:
$$
\((x-y)(x+y)<8\)
$$
2. Analyze the constraints:
- $x$ and $y$ are integers.
- $\(0<y<x\)$ (This means both $x$ and $y$ are positive, and $x$ is greater than $y$ ).
- $\(x+y>3\)$
3. Combine the information:
From step 1, we know $\((x-y)(x+y)<8\)$.
From step 2 , we know $\((x+y)\)$ must be at least 4 (since $x+y>3$ and they are integers).
Also, since $x>y$, the term $\((x-y)\)$ must be at least 1 .
4. Test values for $x$ (starting from the largest choice):
We want the greatest possible value of $x$. Let's test the options:
- $Try \( x=5\)$ :
If $\(x=5\)$, then $y$ must be less than 5 (e.g., $1,2,3,4$ ).
If $\(y=1:(5-1)(5+1)=4 \times 6=24\)$. ( 24 is not $\(<8\)$ )
Since $y$ must be at least 1 , any $x=5$ will make the product too large.
- $Try $\(x=4\)$ :
If $x=4$, the smallest possible $y$ is 1 .
If $\(y=1:(4-1)(4+1)=3 \times 5=15\)$. ( 15 is not $\(<8\)$ )
- $Try \(x=3\)$ :
If $x=3, y$ could be 1 or 2 .
Check $\(y=1:(3-1)(3+1)=2 \times 4=8\)$. ( 8 is not less than 8 )
Check $\(y=2:(3-2)(3+2)=1 \times 5=5\)$.
Does $\(x=3, y=2\)$ satisfy all conditions?
- $\(3^2-2^2=9-4=5<8\)$ (Yes)
- $\(3+2=5>3\)$ (Yes)
- $\(0<2<3\)$ (Yes)
Conclusion:
The greatest integer value for $x$ that satisfies all conditions is $\(\mathbf{3}\)$.
Correct Option:
C. 3
1. Factor the first inequality:
Recall the difference of squares: $\(x^2-y^2=(x-y)(x+y)\)$.
So, the inequality becomes:
$$
\((x-y)(x+y)<8\)
$$
2. Analyze the constraints:
- $x$ and $y$ are integers.
- $\(0<y<x\)$ (This means both $x$ and $y$ are positive, and $x$ is greater than $y$ ).
- $\(x+y>3\)$
3. Combine the information:
From step 1, we know $\((x-y)(x+y)<8\)$.
From step 2 , we know $\((x+y)\)$ must be at least 4 (since $x+y>3$ and they are integers).
Also, since $x>y$, the term $\((x-y)\)$ must be at least 1 .
4. Test values for $x$ (starting from the largest choice):
We want the greatest possible value of $x$. Let's test the options:
- $Try \( x=5\)$ :
If $\(x=5\)$, then $y$ must be less than 5 (e.g., $1,2,3,4$ ).
If $\(y=1:(5-1)(5+1)=4 \times 6=24\)$. ( 24 is not $\(<8\)$ )
Since $y$ must be at least 1 , any $x=5$ will make the product too large.
- $Try $\(x=4\)$ :
If $x=4$, the smallest possible $y$ is 1 .
If $\(y=1:(4-1)(4+1)=3 \times 5=15\)$. ( 15 is not $\(<8\)$ )
- $Try \(x=3\)$ :
If $x=3, y$ could be 1 or 2 .
Check $\(y=1:(3-1)(3+1)=2 \times 4=8\)$. ( 8 is not less than 8 )
Check $\(y=2:(3-2)(3+2)=1 \times 5=5\)$.
Does $\(x=3, y=2\)$ satisfy all conditions?
- $\(3^2-2^2=9-4=5<8\)$ (Yes)
- $\(3+2=5>3\)$ (Yes)
- $\(0<2<3\)$ (Yes)
Conclusion:
The greatest integer value for $x$ that satisfies all conditions is $\(\mathbf{3}\)$.
Correct Option:
C. 3
If x and y are integers in the above inequalities
[#permalink]
03 Mar 2026, 11:33
03 Mar 2026, 11:33
Carcass wrote:
Step-by-Step Solution:
1. Factor the first inequality:
Recall the difference of squares: $\(x^2-y^2=(x-y)(x+y)\)$.
So, the inequality becomes:
$$
\((x-y)(x+y)<8\)
$$
2. Analyze the constraints:
- $x$ and $y$ are integers.
- $\(0<y<x\)$ (This means both $x$ and $y$ are positive, and $x$ is greater than $y$ ).
- $\(x+y>3\)$
3. Combine the information:
You didnt test for x= 4 nd y = 3
From step 1, we know $\((x-y)(x+y)<8\)$.
From step 2 , we know $\((x+y)\)$ must be at least 4 (since $x+y>3$ and they are integers).
Also, since $x>y$, the term $\((x-y)\)$ must be at least 1 .
4. Test values for $x$ (starting from the largest choice):
We want the greatest possible value of $x$. Let's test the options:
- $Try \( x=5\)$ :
If $\(x=5\)$, then $y$ must be less than 5 (e.g., $1,2,3,4$ ).
If $\(y=1:(5-1)(5+1)=4 \times 6=24\)$. ( 24 is not $\(<8\)$ )
Since $y$ must be at least 1 , any $x=5$ will make the product too large.
- $Try $\(x=4\)$ :
If $x=4$, the smallest possible $y$ is 1 .
If $\(y=1:(4-1)(4+1)=3 \times 5=15\)$. ( 15 is not $\(<8\)$ )
- $Try \(x=3\)$ :
If $x=3, y$ could be 1 or 2 .
Check $\(y=1:(3-1)(3+1)=2 \times 4=8\)$. ( 8 is not less than 8 )
Check $\(y=2:(3-2)(3+2)=1 \times 5=5\)$.
Does $\(x=3, y=2\)$ satisfy all conditions?
- $\(3^2-2^2=9-4=5<8\)$ (Yes)
- $\(3+2=5>3\)$ (Yes)
- $\(0<2<3\)$ (Yes)
Conclusion:
The greatest integer value for $x$ that satisfies all conditions is $\(\mathbf{3}\)$.
Correct Option:
C. 3
1. Factor the first inequality:
Recall the difference of squares: $\(x^2-y^2=(x-y)(x+y)\)$.
So, the inequality becomes:
$$
\((x-y)(x+y)<8\)
$$
2. Analyze the constraints:
- $x$ and $y$ are integers.
- $\(0<y<x\)$ (This means both $x$ and $y$ are positive, and $x$ is greater than $y$ ).
- $\(x+y>3\)$
3. Combine the information:
You didnt test for x= 4 nd y = 3
From step 1, we know $\((x-y)(x+y)<8\)$.
From step 2 , we know $\((x+y)\)$ must be at least 4 (since $x+y>3$ and they are integers).
Also, since $x>y$, the term $\((x-y)\)$ must be at least 1 .
4. Test values for $x$ (starting from the largest choice):
We want the greatest possible value of $x$. Let's test the options:
- $Try \( x=5\)$ :
If $\(x=5\)$, then $y$ must be less than 5 (e.g., $1,2,3,4$ ).
If $\(y=1:(5-1)(5+1)=4 \times 6=24\)$. ( 24 is not $\(<8\)$ )
Since $y$ must be at least 1 , any $x=5$ will make the product too large.
- $Try $\(x=4\)$ :
If $x=4$, the smallest possible $y$ is 1 .
If $\(y=1:(4-1)(4+1)=3 \times 5=15\)$. ( 15 is not $\(<8\)$ )
- $Try \(x=3\)$ :
If $x=3, y$ could be 1 or 2 .
Check $\(y=1:(3-1)(3+1)=2 \times 4=8\)$. ( 8 is not less than 8 )
Check $\(y=2:(3-2)(3+2)=1 \times 5=5\)$.
Does $\(x=3, y=2\)$ satisfy all conditions?
- $\(3^2-2^2=9-4=5<8\)$ (Yes)
- $\(3+2=5>3\)$ (Yes)
- $\(0<2<3\)$ (Yes)
Conclusion:
The greatest integer value for $x$ that satisfies all conditions is $\(\mathbf{3}\)$.
Correct Option:
C. 3
You did not test for x=4 and y=3
