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1- Encuentre y grafique en la recta real el conjunto solución de a) \( 1-\frac{8}{7} x \geq 5 \) b) \( 2 x<\frac{x}{4}-3 \) c) \( 2-4 z \leq z+15 \) d) \( -9<1+\frac{1}{4} x \leq \sqrt{7} \) c) \( 2 x \leq 6 x-\frac{4}{3}<8 \) f) \( 5 x>4 x-2>6 x \) g) \( 7 x^{2}+4 x<0 \) h) \( m^{2}+5 m+6 \leq 0 \) d) \( 3 x^{2}-4 x \leq 2 x^{2}-4-x \) k) \( \left(z^{2}-1\right)(z+6) \geq 0 \) i) \( x^{2}>10 x-25 \) 1) \( \frac{t+15}{t-8}>0 \) m) \( \frac{10}{c+1}>-2 \) n) \( \frac{(x-7)\left(x^{2}+x+3\right)}{4 x+5}<0 \) o) \( \frac{24}{u} \geq u \)

Ask by Joseph Chadwick. in Argentina
Jan 23,2025

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Answer

Aquí están las soluciones para cada desigualdad y ecuación, junto con sus intervalos en la recta real: a) \( 1 - \frac{8}{7} x \geq 5 \) \[ x \leq -\frac{7}{2} \] --- b) \( 2 x < \frac{x}{4} - 3 \) \[ x < -\frac{12}{7} \] --- c) \( 2 - 4 z \leq z + 15 \) \[ z \geq -\frac{13}{5} \] --- d) \( -9 < 1 + \frac{1}{4} x \leq \sqrt{7} \) \[ -40 < x \leq 4\sqrt{7} - 4 \] --- e) \( 2 x \leq 6 x - \frac{4}{3} < 8 \) \[ \frac{1}{3} \leq x < \frac{14}{9} \] --- f) \( 5 x > 4 x - 2 > 6 x \) \[ -2 < x < -1 \] --- g) \( 7 x^{2} + 4 x < 0 \) \[ -\frac{4}{7} < x < 0 \] --- h) \( m^{2} + 5 m + 6 \leq 0 \) \[ -3 \leq m \leq -2 \] --- i) \( 3 x^{2} - 4 x \leq 2 x^{2} - 4 - x \) \[ \text{Sin solución} \] --- j) \( (z^{2} - 1)(z + 6) \geq 0 \) \[ z \leq -6 \quad \text{o} \quad z \geq 1 \] --- k) \( x^{2} > 10 x - 25 \) \[ x \neq 5 \] --- l) \( \frac{t + 15}{t - 8} > 0 \) \[ t < -15 \quad \text{o} \quad t > 8 \] --- m) \( \frac{10}{c + 1} > -2 \) \[ c < -6 \quad \text{o} \quad c > -1 \] --- n) \( \frac{(x - 7)(x^{2} + x + 3)}{4 x + 5} < 0 \) \[ -\frac{5}{4} < x < 7 \] --- o) \( \frac{24}{u} \geq u \) \[ u \leq -2\sqrt{6} \quad \text{o} \quad 0 < u \leq 2\sqrt{6} \] Si necesitas más detalles sobre alguna solución, házmelo saber.

Solution

Solve the inequality by following steps: - step0: Solve for \(x\): \(2x<\frac{x}{4}-3\) - step1: Multiply both sides: \(2x\times 4<\left(\frac{x}{4}-3\right)\times 4\) - step2: Multiply the terms: \(8x4 x-2>6 x \). Solve the system of inequalities by following steps: - step0: Solve for \(x\): \(\left\{ \begin{array}{l}5x>4x-2\\4x-2>6x\end{array}\right.\) - step1: Solve the inequality: \(\left\{ \begin{array}{l}x>-2\\x<-1\end{array}\right.\) - step2: Find the intersection: \(-2-40\\x\leq 4\sqrt{7}-4\end{array}\right.\) - step2: Find the intersection: \(-4010 x-25 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}>10x-25\) - step1: Move the expression to the left side: \(x^{2}-\left(10x-25\right)>0\) - step2: Remove the parentheses: \(x^{2}-10x+25>0\) - step3: Rewrite the expression: \(x^{2}-10x+25=0\) - step4: Factor the expression: \(\left(x-5\right)^{2}=0\) - step5: Simplify the expression: \(x-5=0\) - step6: Move the constant to the right side: \(x=0+5\) - step7: Remove 0: \(x=5\) - step8: Determine the test intervals: \(\begin{align}&x<5\\&x>5\end{align}\) - step9: Choose a value: \(\begin{align}&x_{1}=4\\&x_{2}=6\end{align}\) - step10: Test the chosen value: \(\begin{align}&x<5\textrm{ }\textrm{is the solution}\\&x>5\textrm{ }\textrm{is the solution}\end{align}\) - step11: The final solution is \(x\neq 5:\) \(x\neq 5\) Solve the equation \( 3 x^{2}-4 x \leq 2 x^{2}-4-x \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(3x^{2}-4x\leq 2x^{2}-4-x\) - step1: Move the expression to the left side: \(3x^{2}-4x-\left(2x^{2}-4-x\right)\leq 0\) - step2: Subtract the terms: \(x^{2}-3x+4\leq 0\) - step3: Rewrite the expression: \(x^{2}-3x+4=0\) - step4: Add or subtract both sides: \(x^{2}-3x=-4\) - step5: Add the same value to both sides: \(x^{2}-3x+\frac{9}{4}=-4+\frac{9}{4}\) - step6: Simplify the expression: \(\left(x-\frac{3}{2}\right)^{2}=-\frac{7}{4}\) - step7: The statement is false for any value of \(x:\) \(x \notin \mathbb{R}\) - step8: Choose a value: \(x=0\) - step9: Test the chosen value: \(x \in \varnothing \) Solve the equation \( 2 x \leq 6 x-\frac{4}{3}<8 \). Solve the system of inequalities by following steps: - step0: Solve for \(x\): \(\left\{ \begin{array}{l}2x\leq 6x-\frac{4}{3}\\6x-\frac{4}{3}<8\end{array}\right.\) - step1: Solve the inequality: \(\left\{ \begin{array}{l}x\geq \frac{1}{3}\\x<\frac{14}{9}\end{array}\right.\) - step2: Find the intersection: \(\frac{1}{3}\leq x<\frac{14}{9}\) Solve the equation \( 2-4 z \leq z+15 \). Solve the inequality by following steps: - step0: Solve for \(z\): \(2-4z\leq z+15\) - step1: Move the expression to the left side: \(-4z-z\leq 15-2\) - step2: Add and subtract: \(-5z\leq 15-2\) - step3: Add and subtract: \(-5z\leq 13\) - step4: Change the signs: \(5z\geq -13\) - step5: Divide both sides: \(\frac{5z}{5}\geq \frac{-13}{5}\) - step6: Divide the numbers: \(z\geq -\frac{13}{5}\) Solve the equation \( \frac{t+15}{t-8}>0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(\frac{t+15}{t-8}>0\) - step1: Find the domain: \(\frac{t+15}{t-8}>0,t\neq 8\) - step2: Set the numerator and denominator of \(\frac{t+15}{t-8}\) equal to 0\(:\) \(\begin{align}&t+15=0\\&t-8=0\end{align}\) - step3: Calculate: \(\begin{align}&t=-15\\&t=8\end{align}\) - step4: Determine the test intervals: \(\begin{align}&t<-15\\&-158\end{align}\) - step5: Choose a value: \(\begin{align}&t_{1}=-16\\&t_{2}=-4\\&t_{3}=9\end{align}\) - step6: Test the chosen value: \(\begin{align}&t<-15\textrm{ }\textrm{is the solution}\\&-158\textrm{ }\textrm{is the solution}\end{align}\) - step7: The final solution is \(t \in \left(-\infty,-15\right)\cup \left(8,+\infty\right):\) \(t \in \left(-\infty,-15\right)\cup \left(8,+\infty\right)\) - step8: Check if the solution is in the defined range: \(t \in \left(-\infty,-15\right)\cup \left(8,+\infty\right),t\neq 8\) - step9: Find the intersection: \(t \in \left(-\infty,-15\right)\cup \left(8,+\infty\right)\) Solve the equation \( m^{2}+5 m+6 \leq 0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(m^{2}+5m+6\leq 0\) - step1: Rewrite the expression: \(m^{2}+5m+6=0\) - step2: Factor the expression: \(\left(m+2\right)\left(m+3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&m+2=0\\&m+3=0\end{align}\) - step4: Solve the equation: \(\begin{align}&m=-2\\&m=-3\end{align}\) - step5: Determine the test intervals: \(\begin{align}&m<-3\\&-3-2\end{align}\) - step6: Choose a value: \(\begin{align}&m_{1}=-4\\&m_{2}=-\frac{5}{2}\\&m_{3}=-1\end{align}\) - step7: Test the chosen value: \(\begin{align}&m<-3\textrm{ }\textrm{is not a solution}\\&-3-2\textrm{ }\textrm{is not a solution}\end{align}\) - step8: Include the critical value: \(\begin{align}&-3\leq m\leq -2\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(-3\leq m\leq -2:\) \(-3\leq m\leq -2\) Solve the equation \( 7 x^{2}+4 x<0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(7x^{2}+4x<0\) - step1: Rewrite the expression: \(7x^{2}+4x=0\) - step2: Factor the expression: \(x\left(7x+4\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x=0\\&7x+4=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=0\\&x=-\frac{4}{7}\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-\frac{4}{7}\\&-\frac{4}{7}0\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=-\frac{2}{7}\\&x_{3}=1\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-\frac{4}{7}\textrm{ }\textrm{is not a solution}\\&-\frac{4}{7}0\textrm{ }\textrm{is not a solution}\end{align}\) - step8: The final solution is \(-\frac{4}{7}1\end{align}\) - step5: Choose a value: \(\begin{align}&z_{1}=-7\\&z_{2}=-4\\&z_{3}=0\\&z_{4}=2\end{align}\) - step6: Test the chosen value: \(\begin{align}&z<-6\textrm{ }\textrm{is not a solution}\\&-61\textrm{ }\textrm{is the solution}\end{align}\) - step7: Include the critical value: \(\begin{align}&-6\leq z\leq -1\textrm{ }\textrm{is the solution}\\&z\geq 1\textrm{ }\textrm{is the solution}\end{align}\) - step8: The final solution is \(z \in \left[-6,-1\right]\cup \left[1,+\infty\right):\) \(z \in \left[-6,-1\right]\cup \left[1,+\infty\right)\) Solve the equation \( \frac{24}{u} \geq u \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(\frac{24}{u}\geq u\) - step1: Find the domain: \(\frac{24}{u}\geq u ,u\neq 0\) - step2: Move the expression to the left side: \(\frac{24}{u}-u\geq 0\) - step3: Subtract the terms: \(\frac{24-u^{2}}{u}\geq 0\) - step4: Set the numerator and denominator of \(\frac{24-u^{2}}{u}\) equal to 0\(:\) \(\begin{align}&24-u^{2}=0\\&u=0\end{align}\) - step5: Calculate: \(\begin{align}&u=2\sqrt{6}\\&u=-2\sqrt{6}\\&u=0\end{align}\) - step6: Determine the test intervals: \(\begin{align}&u<-2\sqrt{6}\\&-2\sqrt{6}2\sqrt{6}\end{align}\) - step7: Choose a value: \(\begin{align}&u_{1}=-6\\&u_{2}=-2\\&u_{3}=2\\&u_{4}=6\end{align}\) - step8: Test the chosen value: \(\begin{align}&u<-2\sqrt{6}\textrm{ }\textrm{is the solution}\\&-2\sqrt{6}2\sqrt{6}\textrm{ }\textrm{is not a solution}\end{align}\) - step9: Include the critical value: \(\begin{align}&u\leq -2\sqrt{6}\textrm{ }\textrm{is the solution}\\&0-2 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(\frac{10}{c+1}>-2\) - step1: Find the domain: \(\frac{10}{c+1}>-2,c\neq -1\) - step2: Move the expression to the left side: \(\frac{10}{c+1}-\left(-2\right)>0\) - step3: Subtract the terms: \(\frac{12+2c}{c+1}>0\) - step4: Set the numerator and denominator of \(\frac{12+2c}{c+1}\) equal to 0\(:\) \(\begin{align}&12+2c=0\\&c+1=0\end{align}\) - step5: Calculate: \(\begin{align}&c=-6\\&c=-1\end{align}\) - step6: Determine the test intervals: \(\begin{align}&c<-6\\&-6-1\end{align}\) - step7: Choose a value: \(\begin{align}&c_{1}=-7\\&c_{2}=-4\\&c_{3}=0\end{align}\) - step8: Test the chosen value: \(\begin{align}&c<-6\textrm{ }\textrm{is the solution}\\&-6-1\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(c \in \left(-\infty,-6\right)\cup \left(-1,+\infty\right):\) \(c \in \left(-\infty,-6\right)\cup \left(-1,+\infty\right)\) - step10: Check if the solution is in the defined range: \(c \in \left(-\infty,-6\right)\cup \left(-1,+\infty\right),c\neq -1\) - step11: Find the intersection: \(c \in \left(-\infty,-6\right)\cup \left(-1,+\infty\right)\) Solve the equation \( \frac{(x-7)\left(x^{2}+x+3\right)}{4 x+5}<0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(\frac{\left(x-7\right)\left(x^{2}+x+3\right)}{4x+5}<0\) - step1: Find the domain: \(\frac{\left(x-7\right)\left(x^{2}+x+3\right)}{4x+5}<0,x\neq -\frac{5}{4}\) - step2: Set the numerator and denominator of \(\frac{\left(x-7\right)\left(x^{2}+x+3\right)}{4x+5}\) equal to 0\(:\) \(\begin{align}&\left(x-7\right)\left(x^{2}+x+3\right)=0\\&4x+5=0\end{align}\) - step3: Calculate: \(\begin{align}&x=7\\&x=-\frac{5}{4}\end{align}\) - step4: Determine the test intervals: \(\begin{align}&x<-\frac{5}{4}\\&-\frac{5}{4}7\end{align}\) - step5: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=3\\&x_{3}=8\end{align}\) - step6: Test the chosen value: \(\begin{align}&x<-\frac{5}{4}\textrm{ }\textrm{is not a solution}\\&-\frac{5}{4}7\textrm{ }\textrm{is not a solution}\end{align}\) - step7: The final solution is \(-\frac{5}{4} 4 x - 2 > 6 x \) \[ -2 < x < -1 \] **Gráfica:** Se grafica una línea abierta entre \(-2\) y \(-1\). --- g) \( 7 x^{2} + 4 x < 0 \) \[ -\frac{4}{7} < x < 0 \] **Gráfica:** Se grafica una línea abierta entre \(-\frac{4}{7}\) y \(0\). --- h) \( m^{2} + 5 m + 6 \leq 0 \) \[ -3 \leq m \leq -2 \] **Gráfica:** Se grafica una línea sólida entre \(-3\) y \(-2\). --- i) \( 3 x^{2} - 4 x \leq 2 x^{2} - 4 - x \) \[ x \in \varnothing \] **Gráfica:** No hay solución, por lo que no se grafica nada. --- j) \( (z^{2} - 1)(z + 6) \geq 0 \) \[ z \in [-6, -1] \cup [1, +\infty) \] **Gráfica:** Se grafica una línea sólida desde \(-6\) hasta \(-1\) y desde \(1\) hacia la derecha. --- k) \( x^{2} > 10 x - 25 \) \[ x \neq 5 \] **Gráfica:** Se grafica una línea abierta en \(5\) y se extiende hacia ambos lados. --- l) \( \frac{t + 15}{t - 8} > 0 \) \[ t \in (-\infty, -15) \cup (8, +\infty) \] **Gráfica:** Se grafica una línea abierta hacia la izquierda desde \(-15\) y hacia la derecha desde \(8\). --- m) \( \frac{10}{c + 1} > -2 \) \[ c \in (-\infty, -6) \cup (-1, +\infty) \] **Gráfica:** Se grafica una línea abierta hacia la izquierda desde \(-6\) y hacia la derecha desde \(-1\). --- n) \( \frac{(x - 7)(x^{2} + x + 3)}{4 x + 5} < 0 \) \[ -\frac{5}{4} < x < 7 \] **Gráfica:** Se grafica una línea abierta entre \(-\frac{5}{4}\) y \(7\). --- o) \( \frac{24}{u} \geq u \) \[ u \in (-\infty, -2\sqrt{6}] \cup (0, 2\sqrt{6}] \] **Gráfica:** Se grafica una línea sólida hacia la izquierda desde \(-2\sqrt{6}\) y hacia la derecha desde \(0\) hasta \(2\sqrt{6}\). --- Si necesitas más detalles sobre alguna de las soluciones o sus gráficas, házmelo saber.

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Claro, aquí tienes un par de datos interesantes para ampliar tu conocimiento sobre el tema. A lo largo de la historia de las matemáticas, el desarrollo de las desigualdades se remonta a los antiguos griegos, quienes comenzaron a estudiar las propiedades numéricas. Sin embargo, fue en el siglo XIX cuando se formalizaron el concepto y las reglas para manipularlas. Matemáticos como Augustin-Louis Cauchy y David Hilbert contribuyeron significativamente a este campo, proporcionando las bases que todavía utilizamos hoy en día para resolver ecuaciones y desigualdades de maneras sistemáticas. Hoy en día, resolver desigualdades no es solo una práctica académica; también tiene aplicaciones vitales en economía y ciencias sociales. Por ejemplo, al formular restricciones en modelos económicos, se utilizan desigualdades para maximizar beneficios o minimizar costos. También se aplican en algoritmos de optimización, que son fundamentales en la programación y la inteligencia artificial. ¡Las matemáticas son realmente útiles en la vida diaria!

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