Is there a real solution to: $B\le A,B\le C$? Namely, is there $s=(\alpha ,\beta ,\gamma ,\delta )\in {\mathbb{R}}^{4}$ such that $B(s)\le A(s),B(s)\le C(s)$?

Gauge Terrell
2022-07-07
Answered

Is there a real solution to: $B\le A,B\le C$? Namely, is there $s=(\alpha ,\beta ,\gamma ,\delta )\in {\mathbb{R}}^{4}$ such that $B(s)\le A(s),B(s)\le C(s)$?

You can still ask an expert for help

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Graph the solution set for the system of linear inequalities

$x\ge 3$

$y\ge -1$

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Graph the solution set of each given system of linear inequalities.

x < 1

2x + y ≤ 4

asked 2022-07-09

Assume $0<r<1<p$ and $k,\lambda \ge 0$. Let the functions $y,z\in {C}^{1}(0,T)$ satisfy $y\ge 0$, $z>0$ and the system of differential inequalities

${z}^{\prime}\ge {y}^{p}$

${y}^{\prime}+\lambda y+k{z}^{\prime r}\ge z$

on $(0,T)$. Then $T<\infty $ ($T$ is the maximum lifetime of the solution).

${z}^{\prime}\ge {y}^{p}$

${y}^{\prime}+\lambda y+k{z}^{\prime r}\ge z$

on $(0,T)$. Then $T<\infty $ ($T$ is the maximum lifetime of the solution).

asked 2022-07-10

$\{\begin{array}{l}{y}^{2}-3\ge 0\\ 16{y}^{4}-96{y}^{2}\ge 0\end{array}$

the solution for the first inequality is $y\le -\sqrt{3}$ or $y\ge \sqrt{3}$ and the solution for the second inequality is $-\sqrt{6}\le y\le \sqrt{6}$. Then for my result the solution for the system is

Then for my result the solution for the system is or $\sqrt{3}\le y\le \sqrt{6}$

the solution for the first inequality is $y\le -\sqrt{3}$ or $y\ge \sqrt{3}$ and the solution for the second inequality is $-\sqrt{6}\le y\le \sqrt{6}$. Then for my result the solution for the system is

Then for my result the solution for the system is or $\sqrt{3}\le y\le \sqrt{6}$

asked 2022-06-22

How to solve the following system of inequalities rigorously

$\{\begin{array}{l}\mathrm{\Phi}\ge \frac{1}{2}({x}^{2}-{y}^{2}-{z}^{2})+x+y+z-\frac{3}{8},\\ \mathrm{\Phi}\le \frac{1}{2}({x}^{2}-{y}^{2}-{z}^{2})+\frac{1}{2}x+y+z-\frac{9}{8}\\ \mathrm{\Phi}\le \frac{1}{2}({x}^{2}-{y}^{2}-{z}^{2})+\frac{1}{8}\\ y\ge \frac{1}{2},z\ge 1\end{array}$

The task is find the range of $\mathrm{\Phi}$ such that this system of inequalities in $x,y,z$ has solutions in $\mathbb{R}$.

$\{\begin{array}{l}\mathrm{\Phi}\ge \frac{1}{2}({x}^{2}-{y}^{2}-{z}^{2})+x+y+z-\frac{3}{8},\\ \mathrm{\Phi}\le \frac{1}{2}({x}^{2}-{y}^{2}-{z}^{2})+\frac{1}{2}x+y+z-\frac{9}{8}\\ \mathrm{\Phi}\le \frac{1}{2}({x}^{2}-{y}^{2}-{z}^{2})+\frac{1}{8}\\ y\ge \frac{1}{2},z\ge 1\end{array}$

The task is find the range of $\mathrm{\Phi}$ such that this system of inequalities in $x,y,z$ has solutions in $\mathbb{R}$.

asked 2022-06-18

Solve the system of linear inequalities with parameters

$\begin{array}{}\text{(*)}& \{\begin{array}{l}0\le \phantom{-2\phantom{\rule{thickmathspace}{0ex}}}x+2\phantom{\rule{thinmathspace}{0ex}}y-3\phantom{\rule{thinmathspace}{0ex}}b+3\phantom{\rule{thinmathspace}{0ex}}a\le 2\\ 0\le -2\phantom{\rule{thinmathspace}{0ex}}x-3\phantom{\rule{thinmathspace}{0ex}}y+6\phantom{\rule{thinmathspace}{0ex}}b\phantom{\phantom{\rule{thickmathspace}{0ex}}+3a\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}}\le 1\\ 0\le x\le 1\\ 0\le y\le 2\\ 0\le a\le 1\\ 0\le b\le 1\end{array}\end{array}$

Here x,y are unknown variables and a,b are parameters.

My attempt. By adding the inequalities with some coeficients I separated the variables and get the simple system

$\begin{array}{}\text{(**)}& \{\begin{array}{l}0\le y+6a\le 5,\\ 0\le -x+9a+3b\le 8.\end{array}\end{array}$

and I am able to solve it. But the solutions of the last system are not solution of the initial system!

Maple and wolframAlpha cant solve the system.

P.S.1 For $a=\frac{63}{100}$ and $b=\frac{59}{100}$ Maple gives the solutions

$\begin{array}{c}\{x=1,\frac{9}{50}\le y,y\le \frac{11}{25}\},\{x=-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100},\frac{9}{50}<y,y<\frac{11}{25}\},\{\frac{9}{50}<y,x<1,y<\frac{11}{25},-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100}<x\},\{y=\frac{11}{25},\frac{61}{100}\le x,x<1\},\{x=-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100},\frac{11}{25}<y,y<\frac{127}{150}\},\{\frac{11}{25}<y,x<-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},y<\frac{127}{150},-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100}<x\},\{x=-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},\frac{11}{25}<y,y<\frac{127}{150}\},\{x=0,\frac{127}{150}\le y,y\le \frac{47}{50}\},\{y=\frac{127}{150},x\le \frac{14}{75},0<x\},\{0<x,\frac{127}{150}<y,x<-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},y<\frac{47}{50}\},\\ \{x=-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},\frac{127}{150}<y,y<\frac{47}{50}\}\end{array}$

$\begin{array}{}\text{(*)}& \{\begin{array}{l}0\le \phantom{-2\phantom{\rule{thickmathspace}{0ex}}}x+2\phantom{\rule{thinmathspace}{0ex}}y-3\phantom{\rule{thinmathspace}{0ex}}b+3\phantom{\rule{thinmathspace}{0ex}}a\le 2\\ 0\le -2\phantom{\rule{thinmathspace}{0ex}}x-3\phantom{\rule{thinmathspace}{0ex}}y+6\phantom{\rule{thinmathspace}{0ex}}b\phantom{\phantom{\rule{thickmathspace}{0ex}}+3a\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}}\le 1\\ 0\le x\le 1\\ 0\le y\le 2\\ 0\le a\le 1\\ 0\le b\le 1\end{array}\end{array}$

Here x,y are unknown variables and a,b are parameters.

My attempt. By adding the inequalities with some coeficients I separated the variables and get the simple system

$\begin{array}{}\text{(**)}& \{\begin{array}{l}0\le y+6a\le 5,\\ 0\le -x+9a+3b\le 8.\end{array}\end{array}$

and I am able to solve it. But the solutions of the last system are not solution of the initial system!

Maple and wolframAlpha cant solve the system.

P.S.1 For $a=\frac{63}{100}$ and $b=\frac{59}{100}$ Maple gives the solutions

$\begin{array}{c}\{x=1,\frac{9}{50}\le y,y\le \frac{11}{25}\},\{x=-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100},\frac{9}{50}<y,y<\frac{11}{25}\},\{\frac{9}{50}<y,x<1,y<\frac{11}{25},-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100}<x\},\{y=\frac{11}{25},\frac{61}{100}\le x,x<1\},\{x=-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100},\frac{11}{25}<y,y<\frac{127}{150}\},\{\frac{11}{25}<y,x<-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},y<\frac{127}{150},-3/2\phantom{\rule{thinmathspace}{0ex}}y+\frac{127}{100}<x\},\{x=-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},\frac{11}{25}<y,y<\frac{127}{150}\},\{x=0,\frac{127}{150}\le y,y\le \frac{47}{50}\},\{y=\frac{127}{150},x\le \frac{14}{75},0<x\},\{0<x,\frac{127}{150}<y,x<-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},y<\frac{47}{50}\},\\ \{x=-2\phantom{\rule{thinmathspace}{0ex}}y+\frac{47}{25},\frac{127}{150}<y,y<\frac{47}{50}\}\end{array}$

asked 2022-06-24

trying to solve a systems of equations with one inequality

I am trying to create a website that would run off this mathematical formula. I have tried to solve it but I got that there was no answer. I am only in pre-algebra and want a second opinion on if I got the equation right.

$1.2=x-y$

$x>1$

$y>1$

$x>y+2.3$

Is there a solution and if so how do you get it?

I am trying to create a website that would run off this mathematical formula. I have tried to solve it but I got that there was no answer. I am only in pre-algebra and want a second opinion on if I got the equation right.

$1.2=x-y$

$x>1$

$y>1$

$x>y+2.3$

Is there a solution and if so how do you get it?