Pregunta
\( \begin{array}{ll}1.3 & -x^{2}+16>0 \\ 1.4 & \text { Solve for } x \text { and } y \text { simultaneously. } \\ & x-3 y=1 \text { and }(2 x+y-1)(x-y+1)=0 \\ 1.5 & \text { Prove that } \\ & \sqrt{\sqrt[3]{a b}} \cdot \sqrt[b]{b^{3} \sqrt{a}}=a^{\frac{2+b}{6 b}} b^{\frac{b+6}{6 b}}\end{array} \)
Ask by Munoz Simpson. in South Africa
Mar 11,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
- **1.3:** \( -4 < x < 4 \)
- **1.4:** \( (x, y) = (-2, -1) \) and \( \left(\frac{4}{7}, -\frac{1}{7}\right) \)
- **1.5:** The equality is false.
Solución
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
\(-x^{2}+16>0\)
- step1: Rewrite the expression:
\(-x^{2}+16=0\)
- step2: Move the constant to the right side:
\(-x^{2}=0-16\)
- step3: Remove 0:
\(-x^{2}=-16\)
- step4: Change the signs:
\(x^{2}=16\)
- step5: Simplify the expression:
\(x=\pm \sqrt{16}\)
- step6: Simplify:
\(x=\pm 4\)
- step7: Separate into possible cases:
\(\begin{align}&x=4\\&x=-4\end{align}\)
- step8: Determine the test intervals:
\(\begin{align}&x<-4\\&-44\end{align}\)
- step9: Choose a value:
\(\begin{align}&x_{1}=-5\\&x_{2}=0\\&x_{3}=5\end{align}\)
- step10: Test the chosen value:
\(\begin{align}&x<-4\textrm{ }\textrm{is not a solution}\\&-44\textrm{ }\textrm{is not a solution}\end{align}\)
- step11: The final solution is \(-40
\]
**Step 1:** Rearrange the inequality:
\[
-x^{2} + 16 > 0 \implies 16 > x^{2} \implies x^{2} < 16
\]
**Step 2:** Take the square root of both sides:
\[
-\sqrt{16} < x < \sqrt{16} \implies -4 < x < 4
\]
**Final Result for 1.3:**
\[
-4 < x < 4
\]
---
### 1.4 Solve the system of equations:
\[
x - 3y = 1 \quad \text{and} \quad (2x + y - 1)(x - y + 1) = 0
\]
**Step 1:** Solve the first equation for \(x\):
\[
x = 3y + 1
\]
**Step 2:** Substitute \(x\) into the second equation:
1. For \(2x + y - 1 = 0\):
\[
2(3y + 1) + y - 1 = 0 \implies 6y + 2 + y - 1 = 0 \implies 7y + 1 = 0 \implies y = -\frac{1}{7}
\]
Substitute \(y\) back to find \(x\):
\[
x = 3(-\frac{1}{7}) + 1 = -\frac{3}{7} + 1 = \frac{4}{7}
\]
2. For \(x - y + 1 = 0\):
\[
(3y + 1) - y + 1 = 0 \implies 3y - y + 2 = 0 \implies 2y + 2 = 0 \implies y = -1
\]
Substitute \(y\) back to find \(x\):
\[
x = 3(-1) + 1 = -3 + 1 = -2
\]
**Final Result for 1.4:**
\[
(x,y) = (-2,-1) \cup (x,y) = \left(\frac{4}{7}, -\frac{1}{7}\right)
\]
---
### 1.5 Prove the equality:
\[
\sqrt{\sqrt[3]{ab}} \cdot \sqrt[b]{b^{3} \sqrt{a}} = a^{\frac{2+b}{6b}} b^{\frac{b+6}{6b}}
\]
**Result for 1.5:**
The statement is **false**.
---
### Summary of Results:
- **1.3:** \( -4 < x < 4 \)
- **1.4:** \( (x,y) = (-2,-1) \cup (x,y) = \left(\frac{4}{7}, -\frac{1}{7}\right) \)
- **1.5:** The equality is false.
Respondido por UpStudy AI y revisado por un tutor profesional
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Simplifique esta solución Extra Insights
To tackle the first inequality, \(-x^{2}+16>0\), let’s rewrite it as \(x^2 < 16\). This means \(-4 < x < 4\). So, \(x\) lies between -4 and 4! It’s like finding hidden treasure on a number line; you just need to stay within those bounds! For the simultaneous equations \(x-3y=1\) and \((2x+y-1)(x-y+1)=0\), start solving the first one for \(y\): \(y=\frac{x-1}{3}\). Now plug this into either factor from the second equation, starting with \(2x+y-1=0\) to create a new equation. This leads you to coordinate pairs that represent the intersection points. Remember, it’s like putting puzzle pieces together; make sure everything fits before you declare victory!
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