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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