Example Question - integer solutions
Here are examples of questions we've helped users solve.
Solving a System of Inequalities
<p>Решим систему неравенств:</p> <p>\[ \begin{align*} x^2 &< 9 \\ 2x &> 1 \end{align*} \]</p> <p>Решим первое неравенство:</p> <p>\[ x^2 < 9 \Rightarrow -3 < x < 3 \]</p> <p>Решим второе неравенство:</p> <p>\[ 2x > 1 \Rightarrow x > \frac{1}{2} \]</p> <p>Комбинируем оба неравенства:</p> <p>\[ \frac{1}{2} < x < 3 \]</p> <p>Теперь найдём наибольшее целочисленное решение системы неравенств:</p> <p>Наибольшее целое число, которое удовлетворяет условию, это \( x = 2 \).</p>
Solving Integer Inequalities
好的，我们来一步步解这个问题。 首先，我们看第一个不等式 \(x \geq -1\)。由于 \(x\) 是整数，\(x\) 可以是 -1, 0, 1, 2, 等。 接着，我们解第二个不等式 \(3x + 2 < 6\)。 让我们先把不等式中的常数项移到右边去： \(3x < 6 - 2\) 这样就得 \(3x < 4\)。 然后，我们把不等式两边都除以 3（因为我们知道 \(x\) 是整数，所以我们不必担心除法会导致结果不是整数）： \(x < \frac{4}{3}\) 由于 \(x\) 是整数，它不能是分数。因此，\(x\) 必须小于 1（也就是 4/3 的整数部分）。 将两个不等式结合起来，我们得到： \(x \geq -1\) 且 \(x < 1\) 整数 \(x\) 的所有可能值是 \(-1\) 和 \(0\)。 所以，整数 \(x\) 的所有值是 { -1, 0 }。
Solving a System of Equations with Integer Solutions
Sure, let's solve the questions provided in the image one by one. 1. Solve in \( \mathbb{Z} \) the system \( (x^2 - y^2 = 5440, \operatorname{gcd}(x;y) = 8) \) We know that \( x^2 - y^2 = (x+y)(x-y) \), and since the greatest common divisor of \( x \) and \( y \) is 8, we can write \( x = 8m \) and \( y = 8n \), where \( m \) and \( n \) are integers that are coprime (since if \( m \) and \( n \) had a common factor, then \( x \) and \( y \) would have a greater common divisor than 8). Now our equation becomes: \[ (8m + 8n)(8m - 8n) = 5440 \] \[ 64(m + n)(m - n) = 5440 \] \[ (m + n)(m - n) = \frac{5440}{64} \] \[ (m + n)(m - n) = 85 \] The factors of 85 are 1, 5, 17, and 85. Since \( m \) and \( n \) are integers and \( m > n \), the pairs \((m, n)\) that satisfy \( (m+n)(m-n) = 85 \) are \((43, 42)\) and \((9, 8)\). So, plugging these values back into \( x = 8m \) and \( y = 8n \), the solutions are: For \((m, n) = (43, 42)\): \[ x = 8 \cdot 43 = 344 \] \[ y = 8 \cdot 42 = 336 \] For \((m, n) = (9, 8)\): \[ x = 8 \cdot 9 = 72 \] \[ y = 8 \cdot 8 = 64 \] Thus, our solutions in \( \mathbb{Z} \) for the system are \((x, y) = (344, 336)\) and \((x, y) = (72, 64)\). If you need assistance with the other problems in the image, please let me know, and I can help you with those as well.
Solving Exponential Equation $2^x + 3^x = 4^x$
The image you've provided shows an equation: \( 2^x + 3^x = 4^x \). To solve this equation for \( x \), we can try to look for integer solutions by inspection or systematically, assuming \( x \) could be a real number. First, let's examine the behavior of each term as \( x \) increases: - \( 2^x \) and \( 3^x \) both grow at an exponential rate, but \( 3^x \) grows more quickly than \( 2^x \) because it has a larger base. - \( 4^x \) also grows at an exponential rate, even more quickly than both \( 2^x \) and \( 3^x \), since \( 4 = 2^2 \). Given the nature of the equation, we can see that as \( x \) becomes large, \( 4^x \) will dominate the left side of the equation making \( 2^x + 3^x \) less than \( 4^x \). Now, let's consider small values of \( x \): - When \( x = 0 \), \( 2^x + 3^x = 4^x \) because each term equals 1 (any nonzero number to the zeroth power is 1). - When \( x = 1 \), \( 2^x + 3^x = 4^x \) because \( 2 + 3 = 5 \) which is not equal to \( 4 \). - When \( x = 2 \), \( 2^x + 3^x = 4^x \) because \( 4 + 9 = 13 \) which is not equal to \( 16 \). The case when \( x = 0 \) is the only integer solution that satisfies the equation since for \( x > 0 \), the left side of the equation will always be smaller than the right side. Let's try to confirm this by solving the equation algebraically. If we did not find any solutions through inspection or want to consider non-integer real number solutions, we'd have to solve the equation algebraically. However, because this equation is transcendental (involves variable exponents across terms that cannot be easily equated), finding such solutions usually requires numerical methods or advanced mathematical techniques beyond primary algebra. In typical pre-calculus or calculus classes, if this problem was given, the teacher would expect the students to find integer solutions through inspection, as described. For non-integer solutions, students would generally be instructed to use a graphing calculator or numerical methods to approximate solutions, but algebraic methods for finding these approximate solutions without a calculator are largely impractical for this form of the equation. As concluded from inspection, the only solution that can easily be found is \( x = 0 \).
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