How to solve the following pair of equation

1. ${x}^{2}+12x+{y}^{2}-4y=24$

2. ${x}^{2}-6x+{y}^{2}+8y=25$

1. ${x}^{2}+12x+{y}^{2}-4y=24$

2. ${x}^{2}-6x+{y}^{2}+8y=25$

fythynwyrk0
2022-07-10
Answered

How to solve the following pair of equation

1. ${x}^{2}+12x+{y}^{2}-4y=24$

2. ${x}^{2}-6x+{y}^{2}+8y=25$

1. ${x}^{2}+12x+{y}^{2}-4y=24$

2. ${x}^{2}-6x+{y}^{2}+8y=25$

You can still ask an expert for help

Melina Richard

Answered 2022-07-11
Author has **14** answers

Subtract the second equation from the first (collect like terms), get a linear equation, solve for $y,$ substitute back, solve the quadratic for $x,$ substitute back for $y.$

Ximena Skinner

Answered 2022-07-12
Author has **7** answers

Notice that the two equations have the same form as

$1\cdot {x}^{2}+ax+1\cdot {y}^{2}+by=c.$

So, if you subtract one from the other, then you'll get a form $y=dx+e.$.

$1\cdot {x}^{2}+ax+1\cdot {y}^{2}+by=c.$

So, if you subtract one from the other, then you'll get a form $y=dx+e.$.

asked 2022-05-28

Finding equilibrium for differential equations

${x}^{\prime}={x}^{2}+{y}^{2}-5$

${y}^{\prime}={x}^{2}+2{y}^{2}-9$

My goal is to find the equilibrium points.

$(\sqrt{5-{y}^{2}},\pm 2)$, $(\pm \sqrt{{\displaystyle \frac{9}{11}}},\sqrt{5-{x}^{2}})$, $(\sqrt{9-2{y}^{2}},\pm 2)$ and $(\pm \sqrt{{\displaystyle \frac{11}{3}}},{\displaystyle \frac{\sqrt{9-{x}^{2}}}{2}})$

${x}^{\prime}={x}^{2}+{y}^{2}-5$

${y}^{\prime}={x}^{2}+2{y}^{2}-9$

My goal is to find the equilibrium points.

$(\sqrt{5-{y}^{2}},\pm 2)$, $(\pm \sqrt{{\displaystyle \frac{9}{11}}},\sqrt{5-{x}^{2}})$, $(\sqrt{9-2{y}^{2}},\pm 2)$ and $(\pm \sqrt{{\displaystyle \frac{11}{3}}},{\displaystyle \frac{\sqrt{9-{x}^{2}}}{2}})$

asked 2022-07-07

Find basis of solutions of this linear system

Supposed to find basis of the subspace of vector space ${\mathbb{R}}^{3}$ of solutions of this linear system of equations:

$y=\{\begin{array}{l}{x}_{1}+2{x}_{2}-{x}_{3}=0\\ 2{x}_{1}+7{x}_{2}-2{x}_{3}=0\\ -{x}_{1}+3{x}_{2}+{x}_{3}=0\end{array}$

I solve this system and I got: ${x}_{1}={x}_{3}$ and ${x}_{2}=0$

$\overrightarrow{x}=\left[\begin{array}{c}{x}_{1}\\ 0\\ {x}_{1}\end{array}\right]={x}_{1}\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+0\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Is the basis: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?

Supposed to find basis of the subspace of vector space ${\mathbb{R}}^{3}$ of solutions of this linear system of equations:

$y=\{\begin{array}{l}{x}_{1}+2{x}_{2}-{x}_{3}=0\\ 2{x}_{1}+7{x}_{2}-2{x}_{3}=0\\ -{x}_{1}+3{x}_{2}+{x}_{3}=0\end{array}$

I solve this system and I got: ${x}_{1}={x}_{3}$ and ${x}_{2}=0$

$\overrightarrow{x}=\left[\begin{array}{c}{x}_{1}\\ 0\\ {x}_{1}\end{array}\right]={x}_{1}\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+0\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Is the basis: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?

asked 2022-07-15

How to solve this coupled 2nd order Differential equation of a double pendulum- Runge Kutta method

${\theta}_{1}^{\u2033}=\frac{-g(2{m}_{1}+{m}_{2})sin{\theta}_{1}-{m}_{2}gsin({\theta}_{1}-2{\theta}_{2})-2sin({\theta}_{1}-{\theta}_{2}){m}_{2}({\theta}_{2}^{\prime 2}{l}_{2}+{\theta}_{1}^{\prime 2}{l}_{1}cos({\theta}_{1}-{\theta}_{2})}{{l}_{1}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

${\theta}_{2}^{\u2033}=\frac{2sin({\theta}_{1}-{\theta}_{2})({\theta}_{1}^{\prime}{l}_{1}({m}_{1}+{m}_{2})+g({m}_{1}+{m}_{2})cos{\theta}_{1}+{\theta}_{2}^{\prime 2}{l}_{2}{m}_{2}cos({\theta}_{1}-{\theta}_{2}))}{{l}_{2}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method?

${m}_{1},{m}_{2}=$=masses of pendulum 1 and 2, ${\theta}_{1},{\theta}_{2}=$= angles formed by the pendulums, ${\theta}_{1}^{\prime}={\omega}_{1},{\theta}_{2}^{\prime}={\omega}_{2}$

${\theta}_{1}^{\u2033}=\frac{-g(2{m}_{1}+{m}_{2})sin{\theta}_{1}-{m}_{2}gsin({\theta}_{1}-2{\theta}_{2})-2sin({\theta}_{1}-{\theta}_{2}){m}_{2}({\theta}_{2}^{\prime 2}{l}_{2}+{\theta}_{1}^{\prime 2}{l}_{1}cos({\theta}_{1}-{\theta}_{2})}{{l}_{1}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

${\theta}_{2}^{\u2033}=\frac{2sin({\theta}_{1}-{\theta}_{2})({\theta}_{1}^{\prime}{l}_{1}({m}_{1}+{m}_{2})+g({m}_{1}+{m}_{2})cos{\theta}_{1}+{\theta}_{2}^{\prime 2}{l}_{2}{m}_{2}cos({\theta}_{1}-{\theta}_{2}))}{{l}_{2}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method?

${m}_{1},{m}_{2}=$=masses of pendulum 1 and 2, ${\theta}_{1},{\theta}_{2}=$= angles formed by the pendulums, ${\theta}_{1}^{\prime}={\omega}_{1},{\theta}_{2}^{\prime}={\omega}_{2}$

asked 2022-05-19

For how many values of $\alpha $ does this system of equations have infinitely many solutions ?

$\left(\begin{array}{ccc}2& 1& -4\\ 4& 3& -12\\ 1& 2& -8\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}\alpha \\ 5\\ 7\end{array}\right)$

$\left(\begin{array}{ccc}2& 1& -4\\ 4& 3& -12\\ 1& 2& -8\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}\alpha \\ 5\\ 7\end{array}\right)$

asked 2022-07-16

How can I show that there are only finitely many solutions for the following system?

${x}^{2}+yz=x$

${y}^{2}+zx=y$

${z}^{2}+xy=z$

${x}^{2}+yz=x$

${y}^{2}+zx=y$

${z}^{2}+xy=z$

asked 2022-09-05

Consider the coupled nonlinear system of equations given by

$${x}^{3}+{e}^{y}=s\text{}\text{}\text{}\text{}\text{}\text{}\text{}\mathrm{cos}x+xy=t$$

which we wish to be able to solve uniquely for $$(x,y)$$ in terms of $$(s,t)$$. Show this cannot be done at$$(x,y)=(0,0)$$.

$${x}^{3}+{e}^{y}=s\text{}\text{}\text{}\text{}\text{}\text{}\text{}\mathrm{cos}x+xy=t$$

which we wish to be able to solve uniquely for $$(x,y)$$ in terms of $$(s,t)$$. Show this cannot be done at$$(x,y)=(0,0)$$.

asked 2022-05-31

Is it immediately apparent that the solution to the system of equations,

$\begin{array}{}\text{(1)}& \begin{array}{rl}{x}_{1}^{2}& ={x}_{2}+2\\ {x}_{2}^{2}& ={x}_{3}+2\\ {x}_{3}^{2}& ={x}_{4}+2\\ & \vdots \\ {x}_{n}^{2}& ={x}_{1}+2\end{array}\end{array}$

can be given by the roots of unity? Specifically,

$\begin{array}{}\text{(2a)}& x=\frac{{y}_{k}^{2}+1}{{y}_{k}}\end{array}$

where the ${y}_{k}$ are,

$\begin{array}{}\text{(2b)}& \begin{array}{rl}{y}_{k}& =\mathrm{exp}{\textstyle (}\frac{2\pi ik}{{2}^{n}-1}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=0\dots {2}^{n-1}-1\\ {y}_{k}& =\mathrm{exp}{\textstyle (}\frac{2\pi ik}{{2}^{n}+1}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=1\dots {2}^{n-1}\end{array}\end{array}$

Example. Let $n=4$. Then $(1)$ is equivalent to,

$\begin{array}{}\text{(3)}& x=((({x}^{2}-2{)}^{2}-2{)}^{2}-2{)}^{2}-2\end{array}$

Expanded out, $(3)$ is a ${2}^{4}=16$-deg polynomial and its 16 roots are given by $(2)$ where,

${y}_{k}=\mathrm{exp}{\textstyle (}\frac{2\pi ik}{15}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=0\dots 7$

${y}_{k}=\mathrm{exp}{\textstyle (}\frac{2\pi ik}{17}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=1\dots 8$

Ramanujan considered the system $(1)$ for $n=3,4$ in the general case and also as nested radicals. For $x=((({x}^{2}-a{)}^{2}-a{)}^{2}-a{)}^{2}-a$, see this related post. (Interestingly, $n=5$ in the general case is no longer completely solvable in radicals.)

Question:

I observed $(2)$ empirically. How do we prove from first principles that this is indeed the solution?

$\begin{array}{}\text{(1)}& \begin{array}{rl}{x}_{1}^{2}& ={x}_{2}+2\\ {x}_{2}^{2}& ={x}_{3}+2\\ {x}_{3}^{2}& ={x}_{4}+2\\ & \vdots \\ {x}_{n}^{2}& ={x}_{1}+2\end{array}\end{array}$

can be given by the roots of unity? Specifically,

$\begin{array}{}\text{(2a)}& x=\frac{{y}_{k}^{2}+1}{{y}_{k}}\end{array}$

where the ${y}_{k}$ are,

$\begin{array}{}\text{(2b)}& \begin{array}{rl}{y}_{k}& =\mathrm{exp}{\textstyle (}\frac{2\pi ik}{{2}^{n}-1}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=0\dots {2}^{n-1}-1\\ {y}_{k}& =\mathrm{exp}{\textstyle (}\frac{2\pi ik}{{2}^{n}+1}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=1\dots {2}^{n-1}\end{array}\end{array}$

Example. Let $n=4$. Then $(1)$ is equivalent to,

$\begin{array}{}\text{(3)}& x=((({x}^{2}-2{)}^{2}-2{)}^{2}-2{)}^{2}-2\end{array}$

Expanded out, $(3)$ is a ${2}^{4}=16$-deg polynomial and its 16 roots are given by $(2)$ where,

${y}_{k}=\mathrm{exp}{\textstyle (}\frac{2\pi ik}{15}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=0\dots 7$

${y}_{k}=\mathrm{exp}{\textstyle (}\frac{2\pi ik}{17}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=1\dots 8$

Ramanujan considered the system $(1)$ for $n=3,4$ in the general case and also as nested radicals. For $x=((({x}^{2}-a{)}^{2}-a{)}^{2}-a{)}^{2}-a$, see this related post. (Interestingly, $n=5$ in the general case is no longer completely solvable in radicals.)

Question:

I observed $(2)$ empirically. How do we prove from first principles that this is indeed the solution?