I was trying to solve this inequality with two absolute values:

$|2x-3|+7\le 3x-3|x-7|$

I've got an empty set of solutions, but it's not correct.

$|2x-3|+7\le 3x-3|x-7|$

I've got an empty set of solutions, but it's not correct.

enrotlavaec
2022-06-30
Answered

I was trying to solve this inequality with two absolute values:

$|2x-3|+7\le 3x-3|x-7|$

I've got an empty set of solutions, but it's not correct.

$|2x-3|+7\le 3x-3|x-7|$

I've got an empty set of solutions, but it's not correct.

You can still ask an expert for help

Braylon Perez

Answered 2022-07-01
Author has **34** answers

The fastest way for me is to write down LHS-RHS as a piecewise linear function.

$f(x)=|2x-3|+7-3x+3|x-7|\phantom{\rule{0ex}{0ex}}=\{\begin{array}{ll}3-2x+7-3x+3(7-x)=31-8x,& x\u2a7d\frac{3}{2}\\ 2x-3+7-3x+3(7-x)=25-4x,& \frac{3}{2}\u2a7dx\u2a7d7\\ 2x-3+7-3x+3(x-7)=2x-17,& x\u2a7e7\end{array}$

Note that $f(\frac{3}{2})=19,f(7)=-3$

When $x\u2a7d\frac{3}{2},f$ is decreasing so $f(x)>0$;

When $x\u2a7e7,2x-17\u2a7d0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x\u2a7d\frac{17}{2}.$

Therefore $\frac{25}{4}\u2a7dx\u2a7d\frac{17}{2}.$

$f(x)=|2x-3|+7-3x+3|x-7|\phantom{\rule{0ex}{0ex}}=\{\begin{array}{ll}3-2x+7-3x+3(7-x)=31-8x,& x\u2a7d\frac{3}{2}\\ 2x-3+7-3x+3(7-x)=25-4x,& \frac{3}{2}\u2a7dx\u2a7d7\\ 2x-3+7-3x+3(x-7)=2x-17,& x\u2a7e7\end{array}$

Note that $f(\frac{3}{2})=19,f(7)=-3$

When $x\u2a7d\frac{3}{2},f$ is decreasing so $f(x)>0$;

When $x\u2a7e7,2x-17\u2a7d0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x\u2a7d\frac{17}{2}.$

Therefore $\frac{25}{4}\u2a7dx\u2a7d\frac{17}{2}.$

asked 2022-06-24

Given a system of n linear equations. Prove that the system is inconsistent if and only if you can obtain $0=1$, by using linear combinations.

I do not want to apply theorems from linear algebra here. Instead of it I want to use Farkas' lemma.

Every equation can be rewrited in form of two inequalities with $\le $ and $\ge $, now I got $2n$ inequalities instead of $n$ equations.

I think that's the moment for me to apply Farkas' lemma, but why I can always get $0\le 1$ and $0\ge 1$?

I do not want to apply theorems from linear algebra here. Instead of it I want to use Farkas' lemma.

Every equation can be rewrited in form of two inequalities with $\le $ and $\ge $, now I got $2n$ inequalities instead of $n$ equations.

I think that's the moment for me to apply Farkas' lemma, but why I can always get $0\le 1$ and $0\ge 1$?

asked 2022-06-24

Let $({a}_{1},{a}_{2},...,{a}_{n})\in {\mathbb{R}}^{\mathbb{n}}$ and $b\in \mathbb{R}$. Prove that the set:

${F}_{1}:=\{({x}_{1},{x}_{2},...,{x}_{n})\in {\mathbb{R}}^{\mathbb{n}}|\sum _{i=1}^{n}{a}_{i}{x}_{i}\ge b\}$

is closed in ${\mathbb{R}}^{\mathbb{n}}$

${F}_{1}:=\{({x}_{1},{x}_{2},...,{x}_{n})\in {\mathbb{R}}^{\mathbb{n}}|\sum _{i=1}^{n}{a}_{i}{x}_{i}\ge b\}$

is closed in ${\mathbb{R}}^{\mathbb{n}}$

asked 2022-07-15

A nonlinear system

$\left[\begin{array}{c}\dot{{x}_{1}}\\ \dot{{x}_{2}}\end{array}\right]=\left[\begin{array}{c}{{x}_{2}}^{3}+u\\ u\end{array}\right]$

$y={x}_{1}$

as the control objective to make $y$ track ${y}_{d}$, and the tracking error is $e=y-{y}_{d}$.

It is, $\dot{y}=\dot{{x}_{1}}$ and choosing the control law

$u=-{{x}_{2}}^{3}-e+\dot{{y}_{d}}$

gives

$\dot{e}+e=0$

which is stable and converging to zero as the time $t\to \mathrm{\infty}$ (or in other words, the error equation has one pole in -1).

The same control input $u$ applies also to the second equation of the nonlinear system, representing the internal dynamics. With the choice of u as above, $\dot{{x}_{2}}=u$ yields

$\dot{{x}_{2}}+{{x}_{2}}^{3}=\dot{{y}_{d}}-e.$

With the choice of a bounded $\dot{{y}_{d}}$ and e (bounded since $\dot{e}+e=0$), then it is

$|\dot{{y}_{d}}-e|\le D,$

being $D$ a positive constant.

From now on my question starts, as I would like to know what are the math steps to derive the following conclusion:

the example concludes that $|{x}_{2}|\le {D}^{1/3}$(i.e. ${x}_{2}$ is bounded too!), since $\dot{{x}_{2}}<0$ when ${x}_{2}>{D}^{1/3}$, and $\dot{{x}_{2}}>0$ when ${x}_{2}<{D}^{1/3}$. Can someone explain how to derive these inequalities?

$\left[\begin{array}{c}\dot{{x}_{1}}\\ \dot{{x}_{2}}\end{array}\right]=\left[\begin{array}{c}{{x}_{2}}^{3}+u\\ u\end{array}\right]$

$y={x}_{1}$

as the control objective to make $y$ track ${y}_{d}$, and the tracking error is $e=y-{y}_{d}$.

It is, $\dot{y}=\dot{{x}_{1}}$ and choosing the control law

$u=-{{x}_{2}}^{3}-e+\dot{{y}_{d}}$

gives

$\dot{e}+e=0$

which is stable and converging to zero as the time $t\to \mathrm{\infty}$ (or in other words, the error equation has one pole in -1).

The same control input $u$ applies also to the second equation of the nonlinear system, representing the internal dynamics. With the choice of u as above, $\dot{{x}_{2}}=u$ yields

$\dot{{x}_{2}}+{{x}_{2}}^{3}=\dot{{y}_{d}}-e.$

With the choice of a bounded $\dot{{y}_{d}}$ and e (bounded since $\dot{e}+e=0$), then it is

$|\dot{{y}_{d}}-e|\le D,$

being $D$ a positive constant.

From now on my question starts, as I would like to know what are the math steps to derive the following conclusion:

the example concludes that $|{x}_{2}|\le {D}^{1/3}$(i.e. ${x}_{2}$ is bounded too!), since $\dot{{x}_{2}}<0$ when ${x}_{2}>{D}^{1/3}$, and $\dot{{x}_{2}}>0$ when ${x}_{2}<{D}^{1/3}$. Can someone explain how to derive these inequalities?

asked 2022-08-06

Center on the y-axis and touching the line x+y+2=0 at (2, -4)

asked 2020-11-02

Graph the solution set for the of linear inequalities

asked 2022-06-15

$|z-{a}_{k}|\le {R}_{k}$

where $z=x+iy$ (complex number) and ${a}_{k}$ and ${R}_{k}$ are real numbers for $k=1,\dots ,n$. Basically the inequality above shows circle with center ${a}_{k}$ and radius ${R}_{k}$. The question here is, if I write n inequalities as a system of inequalities and then solve this system, the solution will be the intersection of $n$ inequalities. But I want to find the union of $n$ inequalities. Is there any way to do that?

where $z=x+iy$ (complex number) and ${a}_{k}$ and ${R}_{k}$ are real numbers for $k=1,\dots ,n$. Basically the inequality above shows circle with center ${a}_{k}$ and radius ${R}_{k}$. The question here is, if I write n inequalities as a system of inequalities and then solve this system, the solution will be the intersection of $n$ inequalities. But I want to find the union of $n$ inequalities. Is there any way to do that?

asked 2022-06-29

Prove that following systems of inequalities are equivalent.

There are given two systems of inequalities: $b,d,c,e>0$ (first system) and $b+d>0,ec>0,(b+d)(e+c+bd)>be+cd,(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d{)}^{2}ec$ (second system). I need to show that they are equivalent. It is easy to see that first system implies second, because last two inequalities (from the second one) can be transformed to form $bc+{b}^{2}d+de+b{d}^{2}>0$ and $bd((e-c{)}^{2}+{b}^{2}e+bde+bdc+c{d}^{2})>0$ and now it is clear that all inequalities from second system must be true, if $b,c,d,e$ are positive.

There are given two systems of inequalities: $b,d,c,e>0$ (first system) and $b+d>0,ec>0,(b+d)(e+c+bd)>be+cd,(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d{)}^{2}ec$ (second system). I need to show that they are equivalent. It is easy to see that first system implies second, because last two inequalities (from the second one) can be transformed to form $bc+{b}^{2}d+de+b{d}^{2}>0$ and $bd((e-c{)}^{2}+{b}^{2}e+bde+bdc+c{d}^{2})>0$ and now it is clear that all inequalities from second system must be true, if $b,c,d,e$ are positive.