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(25, C2) Tentukan himpunan penyelesaian dari pertaksanaan berikut dan tuliskan solusinya dalam notasi selang: \( \begin{array}{ll}\text { a. }-7<3 x-\frac{1}{2} \leq 10 ; & \text { b. }\left|\frac{2 x-3}{5}-\frac{x}{3}\right|>1 \\ \text { c. } \frac{2 x-5}{x-4} \leq 1\end{array} \)

Ask by Brooks Santiago. in Indonesia
Dec 07,2024

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Answer

a. \( -\frac{13}{6} < x \leq \frac{7}{2} \) b. \( x \in (-\infty,-6) \cup (24,+\infty) \) c. \( 1 \leq x < 4 \)

Solution

Solve the equation \( -7<3x-\frac{1}{2} \leq 10 \). Solve the system of inequalities by following steps: - step0: Solve for \(x\): \(\left\{ \begin{array}{l}-7<3x-\frac{1}{2}\\3x-\frac{1}{2}\leq 10\end{array}\right.\) - step1: Solve the inequality: \(\left\{ \begin{array}{l}x>-\frac{13}{6}\\x\leq \frac{7}{2}\end{array}\right.\) - step2: Find the intersection: \(-\frac{13}{6}1 \). Solve the inequality by following steps: - step0: Solve for \(x\): \(\left|\frac{2x-3}{5}-\frac{x}{3}\right|>1\) - step1: Calculate the absolute value: \(\frac{1}{15}\left|x-9\right|>1\) - step2: Rewrite the expression: \(\frac{\left|x-9\right|}{15}>1\) - step3: Multiply both sides: \(\left|x-9\right|>15\) - step4: Separate into possible cases: \(\begin{align}&x-9>15\\&x-9<-15\end{align}\) - step5: Solve the inequality: \(\begin{align}&x>24\\&x<-6\end{align}\) - step6: Find the union: \(x \in \left(-\infty,-6\right)\cup \left(24,+\infty\right)\) Solve the equation \( \frac{2x-5}{x-4} \leq 1 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(\frac{2x-5}{x-4}\leq 1\) - step1: Find the domain: \(\frac{2x-5}{x-4}\leq 1,x\neq 4\) - step2: Move the expression to the left side: \(\frac{2x-5}{x-4}-1\leq 0\) - step3: Subtract the terms: \(\frac{x-1}{x-4}\leq 0\) - step4: Set the numerator and denominator of \(\frac{x-1}{x-4}\) equal to 0\(:\) \(\begin{align}&x-1=0\\&x-4=0\end{align}\) - step5: Calculate: \(\begin{align}&x=1\\&x=4\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<1\\&14\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=0\\&x_{2}=3\\&x_{3}=5\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<1\textrm{ }\textrm{is not a solution}\\&14\textrm{ }\textrm{is not a solution}\end{align}\) - step9: Include the critical value: \(\begin{align}&1\leq x<4\textrm{ }\textrm{is the solution}\end{align}\) - step10: The final solution is \(1\leq x<4:\) \(1\leq x<4\) - step11: Check if the solution is in the defined range: \(1\leq x<4,x\neq 4\) - step12: Find the intersection: \(1\leq x<4\) a. Solusi dari pertaksanaan \( -7<3x-\frac{1}{2} \leq 10 \) adalah \( -\frac{13}{6} < x \leq \frac{7}{2} \). b. Solusi dari pertaksanaan \( \left|\frac{2x-3}{5}-\frac{x}{3}\right|>1 \) adalah \( x \in (-\infty,-6) \cup (24,+\infty) \). c. Solusi dari pertaksanaan \( \frac{2x-5}{x-4} \leq 1 \) adalah \( 1 \leq x < 4 \).

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Mari kita selesaikan setiap pertaksaan satu per satu! a. Pertaksaan pertama adalah \( -7 < 3x - \frac{1}{2} \leq 10 \). Langkah pertama adalah memecahkan kedua bagian secara terpisah. 1. Untuk bagian pertama \( -7 < 3x - \frac{1}{2} \): - Tambahkan \( \frac{1}{2} \) ke kedua sisi: \(-7 + \frac{1}{2} < 3x\) - Jadi, \( -\frac{14}{2} + \frac{1}{2} < 3x \) dan mendapatkan \( -\frac{13}{2} < 3x \). - Bagilah dengan 3: \( -\frac{13}{6} < x \). 2. Untuk bagian kedua \( 3x - \frac{1}{2} \leq 10 \): - Tambahkan \( \frac{1}{2} \) ke kedua sisi: \( 3x \leq 10 + \frac{1}{2} \) - Jadi, \( 3x \leq \frac{20}{2} + \frac{1}{2} = \frac{21}{2} \). - Bagilah dengan 3: \( x \leq \frac{7}{2} \). Gabungkan kedua hasil tersebut: \( -\frac{13}{6} < x \leq \frac{7}{2} \) Notasi selangnya adalah \( \left(-\frac{13}{6}, \frac{7}{2}\right] \). b. Pertaksaan kedua adalah \( \left|\frac{2x-3}{5}-\frac{x}{3}\right| > 1 \). Langkah pertama adalah menyelesaikan didalam tanda mutlak: 1. Mulai dengan menyamakan penyebut untuk \(\frac{2x-3}{5} - \frac{x}{3}\): - Penyebut bersama adalah 15: \(\frac{3(2x-3) - 5x}{15} = \frac{6x - 9 - 5x}{15} = \frac{x - 9}{15}\). 2. Kemudian, kita punya \(\left|\frac{x - 9}{15}\right| > 1\). - Ini berarti, ada dua kasus: - \( \frac{x - 9}{15} > 1 \) dan \( \frac{x - 9}{15} < -1 \). **Kasus 1:** \( x - 9 > 15 \) sehingga \( x > 24 \). **Kasus 2:** \( x - 9 < -15 \) sehingga \( x < -6 \). Gabungkan solusi: \( x < -6 \) atau \( x > 24 \). Notasi selang untuk solusi ini adalah \( (-\infty, -6) \cup (24, \infty) \). c. Pertaksaan ketiga adalah \( \frac{2x-5}{x-4} \leq 1 \). Langkah pertama adalah menyamakan kedua sisi: 1. Hilangkan 1 dengan mengubahnya ke pecahan: \(\frac{2x-5}{x-4} - 1 \leq 0\) - Ini menjadi: \(\frac{2x - 5 - (x - 4)}{x - 4} \leq 0\) - Menjadi: \(\frac{2x - 5 - x + 4}{x - 4} \leq 0\) - Simplifikasi menjadi: \(\frac{x - 1}{x - 4} \leq 0\). 2. Menentukan titik-titik kritis: - \( x - 1 = 0 \) memberi kita \( x = 1 \). - \( x - 4 = 0 \) memberi kita \( x = 4 \). 3. Melakukan uji tanda pada interval yang dihasilkan oleh \( x = 1 \) dan \( x = 4 \): - Untuk \( x < 1 \), pilih \( x = 0 \): \(\frac{0 - 1}{0 - 4} = \frac{-1}{-4} > 0\). - Untuk \( 1 < x < 4 \), pilih \( x = 2 \): \(\frac{2 - 1}{2 - 4} = \frac{1}{-2} < 0

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