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1. (25, C2) Tentukan himpunan penyelesaian dari pertaksamaan 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 Rogers Cummings. 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: \(1\leq x<4\) Penyelesaian dari pertaksamaan-pertaksamaan tersebut adalah sebagai berikut: a. \( -\frac{13}{6} < x \leq \frac{7}{2} \) b. \( x \in (-\infty, -6) \cup (24, +\infty) \) c. \( 1 \leq x < 4 \)

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Mari kita selesaikan pertaksamaan satu per satu. **a.** Pertaksamaan yang diberikan adalah \(-7 < 3x - \frac{1}{2} \leq 10\). 1. Untuk bagian kiri \(-7 < 3x - \frac{1}{2}\): \[ -7 + \frac{1}{2} < 3x \Rightarrow -\frac{14}{2} + \frac{1}{2} < 3x \Rightarrow -\frac{13}{2} < 3x \Rightarrow x > -\frac{13}{6} \] 2. Untuk bagian kanan \(3x - \frac{1}{2} \leq 10\): \[ 3x \leq 10 + \frac{1}{2} \Rightarrow 3x \leq \frac{20}{2} + \frac{1}{2} \Rightarrow 3x \leq \frac{21}{2} \Rightarrow x \leq \frac{21}{6} \] Sehingga, himpunan penyelesaian untuk a adalah: \[ x \in \left( -\frac{13}{6}, \frac{21}{6} \right] \] **b.** Pertaksamaan yang diberikan adalah \(\left|\frac{2x - 3}{5} - \frac{x}{3}\right| > 1\). 1. Pertama, kita ubah menjadi dua pertaksamaan: \[ \frac{2x - 3}{5} - \frac{x}{3} > 1 \quad \text{atau} \quad \frac{2x - 3}{5} - \frac{x}{3} < -1 \] 2. Untuk \(\frac{2x - 3}{5} - \frac{x}{3} > 1\): \[ \frac{2x - 3}{5} > 1 + \frac{x}{3} \quad \Rightarrow \quad \text{Mencari penyelesaian bagi x} \] Mengalikan seluruh persamaan x dengan kelipatan persekutuan 15, kita dapat melihat bahwa: \[ 6(2x - 3) > 15 + 5x \Rightarrow 12x - 18 > 15 + 5x \Rightarrow 7x > 33 \Rightarrow x > \frac{33}{7} \] 3. Untuk \(\frac{2x - 3}{5} - \frac{x}{3} < -1\): \[ \frac{2x - 3}{5} < -1 + \frac{x}{3} \quad \Rightarrow \quad \text{Mencari penyelesaian bagi x yang sama seperti di atas} \] Mengalikan lagi, kita peroleh: \[ 6(2x - 3) < -15 + 5x \Rightarrow 12x - 18 < -15 + 5x \Rightarrow 7x < 3 \Rightarrow x < \frac{3}{7} \] Jadi, himpunan penyelesaian untuk b adalah: \[ x \in \left( -\infty, \frac{3}{7} \right) \cup \left( \frac{33}{7}, \infty \right) \] **c.** Pertaksamaan yang diberikan adalah \(\frac{2x - 5}{x - 4} \leq 1\). 1. Langkah pertama: \[ \frac{2x - 5}{x - 4} - 1 \leq 0 \Rightarrow \frac{2x - 5 - (x - 4)}{x - 4} \leq 0 \Rightarrow \frac{x - 1}{x - 4} \leq 0 \] 2. Untuk menyelesaikan \(\frac{x - 1}{x - 4} \leq 0\), kita perlu menentukan titik potong: - Titik potong numerik \(x - 1 = 0 \Rightarrow x = 1\) - Titik potong penyebut \(x - 4 = 0 \Rightarrow x = 4\) 3. Uji interval: - Untuk interval \((- \infty, 1)\): positif - Untuk interval \((1, 4)\): negatif - Untuk interval \((4, +\infty)\):

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