 # Systems of Equations Questions and Answers

Recent questions in Systems of equations Despiniosnt 2022-05-24 Answered

### $EXP=\left\{\begin{array}{ll}{n}^{3}\left(\frac{⌊\frac{n+1}{3}⌋+24}{50}\right)& n\le 15\\ {n}^{3}\left(\frac{n+14}{50}\right)& 15\le n\le 36\\ {n}^{3}\left(\frac{⌊\frac{n}{2}⌋+32}{50}\right)& 36\le n\le 100\end{array}$In this equation there are 3 systems of equations, but there are inequalities next to them. Does this mean that for example we use the top equation when$n\le 15\phantom{\rule{0ex}{0ex}}$and so on? Alani Conner 2022-05-23 Answered

### For which value(s) of parameter m is there a solution for this system$\left\{\begin{array}{l}mx+y=m\\ mx+2y=1\\ 2x+my=m+1\end{array}$when does this system of equations have a solution? Case Nixon 2022-05-23 Answered

### Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously1. $x+y=1-z$and2. ${x}^{3}+{y}^{3}=1-{z}^{2}$Find all such $x$, $y$ and $z$. rs450nigglix2 2022-05-23 Answered

### Simultaneous equation difficulty minus$\left(1\right)×3$$\left(3\right)-\left(2\right)$$21x=-7$$3x=-1$ Brooke Webb 2022-05-23 Answered

### Is the following equation regarded as a linear equation?$0{x}_{1}+0{x}_{2}+0{x}_{3}=5$The original question is as below:Solve the linear system given by the following augmented matrix:$\left(\begin{array}{cccc}2& 2& 3& 1\\ 2& 5& 3& 0\\ 0& 0& 0& 5\end{array}\right)$Note the words linear system in the original question. So, I was asking myself whether $0{x}_{1}+0{x}_{2}+0{x}_{3}=5$ is a linear equation. Can we call all of the equations given by the matrix collectively as a linear system? Brennen Fisher 2022-05-22 Answered

### choosing $h$and $k$ such that this system:$\left\{\begin{array}{l}{x}_{1}+h{x}_{2}=2\\ 4{x}_{1}+8{x}_{2}=k\end{array}$Has (a) no solution, (b) a unique solution, and (c) many solutions.First i made an augmented matrix, then performed row reduction:$\left[\begin{array}{ccc}1& h& 2\\ 4& 8& k\end{array}\right]\sim \left[\begin{array}{ccc}1& h& 2\\ 0& 8-4h& k-8\end{array}\right]$Continuing row reduction, i get:$\sim \left[\begin{array}{ccc}1& 0& \frac{k-8}{2\left(h-2\right)}+\frac{k}{4}\\ 0& 1& \frac{k-8}{8-4h}\end{array}\right]$how to go about solving the problem with the matrix i end up with? Rachel Villa 2022-05-22 Answered

### Consider the differential system$\left\{\begin{array}{ll}& {y}^{\prime }\left(t\right)=ay\left(t{\right)}^{3}+bz\left(t\right)\\ & {z}^{\prime }\left(t\right)=cz\left(t{\right)}^{5}-by\left(t\right)\end{array}$with $t>0$$y\left(0\right)={y}_{0},z\left(0\right)={z}_{0},\phantom{\rule{1em}{0ex}}a<0,\phantom{\rule{1em}{0ex}}c<0,\phantom{\rule{1em}{0ex}}b\in \mathbb{R}$the question is to prouve that this system admits a unique solution on $\left[0,+\mathrm{\infty }\right]$? wanaopatays 2022-05-22 Answered

### how to resolve this system of differential equations of order 1?$\left\{\begin{array}{ccc}{\stackrel{˙}{p}}_{1}& =& \frac{1}{z}{p}_{2}{p}_{3}\\ {\stackrel{˙}{p}}_{2}& =& -\frac{1}{z}{p}_{1}{p}_{3}\\ {\stackrel{˙}{p}}_{3}& =& \left(\frac{1}{y}-\frac{1}{x}\right){p}_{1}{p}_{2}\end{array}$where ${p}_{1}\left(0\right)=a,{p}_{2}\left(0\right)=b,{p}_{3}\left(0\right)=c$ and $x,y,z$ are constants.1) how to resolve this system differential equations by hand?2) what if $x=y=M$ where $M$ is a constant? Andy Erickson 2022-05-22 Answered

### Dynamical system equilibrium point increment$\frac{d{x}_{1}}{dx}\left(t\right)={x}_{2}\left(t\right)$$\frac{d{x}_{2}}{dx}\left(t\right)=\frac{-k}{m}{x}_{1}\left(t\right)+\frac{f\left(t\right)}{m}$Then we find the equilibrium point by setting$0={x}_{2}\left(t\right)$$0=\frac{-k}{m}{x}_{1}\left(t\right)+\frac{f\left(t\right)}{m}$Which gives as result${x}_{eq}=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}f/k\\ 0\end{array}\right]$Finally, the book defines the increment with respect to the equilibrium point $\mathrm{\Delta }x\left(t\right)=x\left(t\right)-{x}_{eq}$. Substracting equation $\left(1,3\right)$ and $\left(2,4\right)$ the result is$\frac{d\mathrm{\Delta }{x}_{1}}{dx}\left(t\right)=\mathrm{\Delta }{x}_{2}\left(t\right)$$\frac{d\mathrm{\Delta }{x}_{2}}{dx}\left(t\right)=\frac{-k}{m}\mathrm{\Delta }{x}_{1}\left(t\right)$So far so good, but for the next step they get "the general solution, parametrized by the initial state" as$\mathrm{\Delta }{x}_{1}\left(t\right)=\mathrm{\Delta }{x}_{1}\left(0\right)cos\left(\omega t\right)+\frac{\mathrm{\Delta }{x}_{2}\left(0\right)}{\omega }sin\left(\omega t\right)$$\mathrm{\Delta }{x}_{2}\left(t\right)=-\mathrm{\Delta }{x}_{1}\left(0\right)\omega sin\left(\omega t\right)+\mathrm{\Delta }{x}_{2}\left(0\right)cos\left(\omega t\right)$This result I don't understand where does it comes from, could someone give me a hint? Eliaszowyr1 2022-05-22 Answered

### I have the following linear system:$\begin{array}{rl}& x+y+2z+2=0\\ & 3x-y+14z-6=0\\ & x+2y+5=0\end{array}$I immediately noticed that there was no $z$ term in the last equation and thus determined that I will end with $0z=\text{some number}$ number and therefore, came to the conclusion that there is no solution to the linear system. To see if I was right, I checked with echelon form and that also suggested that there was no solution.$\begin{array}{rl}& {L}_{1}:x+y+2z+2=0\\ & {L}_{2}:3x-y+14z-6=0\\ & {L}_{3}:x+2y+5=0\end{array}$Then $-3{L}_{1}+{L}_{2}\to {L}_{2}$$\begin{array}{rl}& x+y+2z+2=0\\ & 0x+3y-42z-18=0\\ & x+2y+0z+5=0\end{array}$Then $-{L}_{1}+{L}_{3}\to {L}_{3}$$\begin{array}{rl}& x+y+2z+2=0\\ & 0x+3y-42z-18=0\\ & 0x+1y+0z+3=0\end{array}$Then $-{L}_{2}/3+{L}_{3}\to {L}_{3}$$\begin{array}{rl}& x+y+2z+2=0\\ & 0x+3y-42z-18=0\\ & 0x+0y+0z+3=0\end{array}$Firstly, is my answer correct? Despiniosnt 2022-05-21 Answered

### Solving a set of recurrence relations${A}_{n}={B}_{n-1}+{C}_{n-1}$${B}_{n}={A}_{n}+{C}_{n-1}$${C}_{n}={B}_{n}+{C}_{n-1}$${D}_{n}={E}_{n-1}+{G}_{n-1}$${E}_{n}={D}_{n}+{F}_{n-1}$${F}_{n}={G}_{n}+{C}_{n}$${G}_{n}={E}_{n}+{F}_{n-1}$which I would like to solve, with the goal of eventually finding an explicit form of ${E}_{n}$. I started out by looking at only ${A}_{n}$, ${B}_{n}$ and ${C}_{n}$, and found a formula for ${A}_{n}$.${A}_{n}=1/3\sqrt{3}\left(2+\sqrt{3}{\right)}^{n}-1/3\sqrt{3}\left(2-\sqrt{3}{\right)}^{n}$but I cant seem to find the right trick this time. Nylah Burnett 2022-05-21 Answered

### System of ODEs with productsHow can we solve the system of differential equations$\frac{df\left(t\right)}{dt}=-f\left(t\right)h\left(t\right),\frac{dg\left(t\right)}{dt}=-g\left(t\right)h\left(t\right),\frac{dh\left(t\right)}{dt}=1-\left(h\left(t\right){\right)}^{2}$The system does not fall to standard ODE methods. Isaiah Owens 2022-05-21 Answered

### solving equations by the method of elimination$\frac{a}{x}+\frac{b}{y}=\frac{a}{2}+\frac{b}{3}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\dots \left(i\right)$$x+1=y\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\dots \left(ii\right)$We have to solve for x and y, only this time using the method of elimination.From equation $\left(ii\right)$, we get,$\frac{1}{x+1}=\frac{1}{y}⇒\frac{b}{x+1}=\frac{b}{y}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\dots \left(iii\right)$Subtracting $\left(iii\right)$ from $\left(i\right)$, we get,$\frac{a}{x}+\frac{b}{y}-\frac{b}{y}=\frac{a}{2}+\frac{b}{3}-\frac{b}{x+1}$$⇒\frac{a}{x}=\frac{a}{2}+\frac{b}{3}-\frac{b}{x+1}$After that,I really cannot find anything to do.I have taken quite a few other routes, but have hit nothing but dead ends. At this a point a little hint will be appreciated. dglennuo 2022-05-21 Answered

### $M=1+\frac{a}{b}$$S=a+b$So, I put it up like this:$1+\frac{a}{b}=a+b$If $M=S$, how to isolate $a$? Kiana Harper 2022-05-21 Answered

### Find m to the equation:$\left\{\begin{array}{l}2{x}^{3}-\left(y+2\right){x}^{2}+xy=m\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(1\right)\\ {x}^{2}+x-y=1-2m\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(2\right)\end{array}$My try: From $\left(1\right)$ and $\left(2\right)\phantom{\rule{thinmathspace}{0ex}}⇒$:$4{x}^{3}-2\left(y+2\right){x}^{2}+2xy+{x}^{2}+x-y=1\phantom{\rule{0ex}{0ex}}⇔4{x}^{3}-3{x}^{2}+x-1=y\left(2{x}^{2}-2x+1\right)\phantom{\rule{0ex}{0ex}}⇔y=\frac{4{x}^{3}-3{x}^{2}+x-1}{2{x}^{2}-2x+1}\phantom{\rule{0ex}{0ex}}⇔y=2x+\frac{1}{2}-\frac{3}{2\left(2{x}^{2}-2x+1\right)}$From $\left(2\right)\phantom{\rule{thinmathspace}{0ex}}⇒$$2m=1+y-{x}^{2}-x\phantom{\rule{0ex}{0ex}}=1+2x+\frac{1}{2}-\frac{3}{2\left(2{x}^{2}-2x+1\right)}-{x}^{2}-x\phantom{\rule{0ex}{0ex}}=-{x}^{2}+x+\frac{3}{2}-\frac{3}{2\left(2{x}^{2}-2x+1\right)}$And I don't know how to contine,The result is: $m\le 1-\frac{\sqrt{3}}{2}$ Shamar Reese 2022-05-21 Answered

### Solve following series of equations ($n+2$ equations $n+2$ variables):${k}_{0}{q}_{0}+\lambda {q}_{0}+{c}_{0}=0,\phantom{\rule{0ex}{0ex}}{k}_{1}{q}_{1}+\lambda {q}_{1}+{c}_{1}=0,\phantom{\rule{0ex}{0ex}}{k}_{n}{q}_{n}+\lambda {q}_{n}+{c}_{n}=0,\phantom{\rule{0ex}{0ex}}{q}_{1}+{q}_{2}+....+{q}_{n}=1.$The variables are ${q}_{0},{q}_{1},.....,{q}_{n}$ and $\lambda$. Note that $k$ and $c$ are series of constants. Trevor Wood 2022-05-21 Answered

### Applications of polynomial systems of equationsWhat are some applications of Polynomial Systems of Equations? Marianna Stone 2022-05-21 Answered

### How to solve coupled linear 1st order PDEIt is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if${\mathrm{\partial }}_{t}f+a{\mathrm{\partial }}_{x}f=bf$we have that $\frac{df}{dt}=bf$ on the characteristic curve of $\frac{dx}{dt}=a$. From this we deduce that $f\left(t,x\right)=g\left(C\right){e}^{bt}$ where $x=at+C$.Now, how does this work when $f$ is multidimensional. Can I solve equations on the following form by characteristics, or by any other means?${\mathrm{\partial }}_{t}{f}_{i}\left(t,x\right)+\sum _{j}{A}_{ij}{\mathrm{\partial }}_{x}{f}_{j}\left(t,x\right)=\sum _{j}{B}_{ij}{f}_{j}\left(t,x\right)$where the components of $A$ and $B$ might be dependent on $x$ and $t$.In particular, I am trying to solve the following,$\left\{\begin{array}{l}{\mathrm{\partial }}_{t}f+\frac{c}{t}{\mathrm{\partial }}_{x}g=-\left(a+\frac{1}{t}\right)f\\ {\mathrm{\partial }}_{t}g+\frac{c}{t}{\mathrm{\partial }}_{x}f=-\left(b+\frac{1}{t}\right)g\end{array}$where $f$ and $g$ are functions of $x$ and $t$ , where $t>{t}_{0}>0$, $c\ne 0$. Any help is highly appreciated. Davin Fields 2022-05-21 Answered

### I want to show that the following system of equations does not have a solution, but I do not know how to do this${w}_{1}+{w}_{2}=\frac{1}{2}$${w}_{1}{s}_{1}+{w}_{2}{s}_{2}=\frac{1}{6}$${w}_{1}{t}_{1}+{w}_{2}{t}_{2}=\frac{1}{6}$${w}_{1}{s}_{1}{t}_{1}+{w}_{2}{s}_{2}{t}_{2}=\frac{1}{24}$${w}_{1}{s}_{1}^{2}+{w}_{2}{s}_{2}^{2}=\frac{1}{12}$${w}_{1}{t}_{1}^{2}+{w}_{2}{t}_{2}^{2}=\frac{1}{12}$that all ${t}_{1},{t}_{2},{w}_{1},{w}_{2},{s}_{1},{s}_{2}$ are unknown. velitshh 2022-05-21 Answered

### system of equations:$\sqrt{x}+y=7$$\sqrt{y}+x=11$Its pretty visible that the solution is $\left(x,y\right)=\left(9,4\right)$For this, I put $x={p}^{2}$ and $y={q}^{2}$. Then I subtracted one equation from the another such that I got $4$ on RHS and factorized LHS to get two factors in terms of $p$ and $q$.Then $4$ can be represented as $2\ast 2$, $4\ast 1$ or $1\ast 4$. Comparing the two factors on both sides, I got the solution.As you can see, the major drawback here is that I assumed this system has only integral solutions and then went further. Is there any way I can prove that this system indeed has only integral solutions or is there any other elegant way to solve this question?

In simple terms, systems of equations represent a special set of simultaneous equations where the equation system is used as a finite element. The trick here is to find common solutions, which is exactly what systems of equations solver must achieve. If this does not sound clear to you, take a look at some systems of equations answers below and see those with explanations. The solution will always come in three variables (namely, x, y, and z), which will represent your ordered triple. See systems of equations solutions for more examples of how it works in practice.