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High School
Algebra
Algebra I
Systems of equations

Recent questions in Systems of equations

Andy Erickson 2022-05-22

Dynamical system equilibrium point increment
$\frac{d x_{1}}{d x} (t) = x_{2} (t)$
$\frac{d x_{2}}{d x} (t) = \frac{- k}{m} x_{1} (t) + \frac{f (t)}{m}$
Then we find the equilibrium point by setting
$0 = x_{2} (t)$
$0 = \frac{- k}{m} x_{1} (t) + \frac{f (t)}{m}$
Which gives as result
$x_{e q} = [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} f / k \\ 0 \end{matrix}]$
Finally, the book defines the increment with respect to the equilibrium point $Δ x (t) = x (t) - x_{e q}$ . Substracting equation $(1, 3)$ and $(2, 4)$ the result is
$\frac{d Δ x_{1}}{d x} (t) = Δ x_{2} (t)$
$\frac{d Δ x_{2}}{d x} (t) = \frac{- k}{m} Δ x_{1} (t)$
So far so good, but for the next step they get "the general solution, parametrized by the initial state" as
$Δ x_{1} (t) = Δ x_{1} (0) c o s (ω t) + \frac{Δ x_{2} (0)}{ω} s i n (ω t)$
$Δ x_{2} (t) = - Δ x_{1} (0) ω s i n (ω t) + Δ x_{2} (0) c o s (ω t)$
This result I don't understand where does it comes from, could someone give me a hint?

Eliaszowyr1 2022-05-22

I have the following linear system:
$\begin{aligned} x + y + 2 z + 2 = 0 \\ 3 x - y + 14 z - 6 = 0 \\ x + 2 y + 5 = 0 \end{aligned}$
I immediately noticed that there was no $z$ term in the last equation and thus determined that I will end with $0 z = some number$ 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.
$\begin{aligned} L_{1} : x + y + 2 z + 2 = 0 \\ L_{2} : 3 x - y + 14 z - 6 = 0 \\ L_{3} : x + 2 y + 5 = 0 \end{aligned}$
Then $- 3 L_{1} + L_{2} \to L_{2}$
$\begin{aligned} x + y + 2 z + 2 = 0 \\ 0 x + 3 y - 42 z - 18 = 0 \\ x + 2 y + 0 z + 5 = 0 \end{aligned}$
Then $- L_{1} + L_{3} \to L_{3}$
$\begin{aligned} x + y + 2 z + 2 = 0 \\ 0 x + 3 y - 42 z - 18 = 0 \\ 0 x + 1 y + 0 z + 3 = 0 \end{aligned}$
Then $- L_{2} / 3 + L_{3} \to L_{3}$
$\begin{aligned} x + y + 2 z + 2 = 0 \\ 0 x + 3 y - 42 z - 18 = 0 \\ 0 x + 0 y + 0 z + 3 = 0 \end{aligned}$
Firstly, is my answer correct?

Simone Werner 2022-05-22

I have the following system of equations where the m's are known but $a, b, c, x, y, z$ are unknown. How does one go about solving this system? All the usual linear algebra tricks I know don't apply and I don't want to do it through tedious substitutions.
$\begin{aligned} a + b + c & = m_{0} \\ a x + b y + c z & = m_{1} \\ a x^{2} + b y^{2} + c z^{2} & = m_{2} \\ a x^{3} + b y^{3} + c z^{3} & = m_{3} \\ a x^{4} + b y^{4} + c z^{4} & = m_{4} \\ a x^{5} + b y^{5} + c z^{5} & = m_{5} \end{aligned}$

Elisha Kelly 2022-05-22

Solve the system of equations:
$\begin{matrix} 2 x^{5} - 2 x^{3} y - x y^{2} + 10 x^{3} + y^{2} - 5 y = 0 \\ (x + 1) \sqrt{y + 5} = y - 3 x^{2} + x - 2 \end{matrix}$

Despiniosnt 2022-05-21

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 $A_{n}$ .
$A_{n} = 1 / 3 \sqrt{3} (2 + \sqrt{3})^{n} - 1 / 3 \sqrt{3} (2 - \sqrt{3})^{n}$
but I cant seem to find the right trick this time.

Nylah Burnett 2022-05-21

System of ODEs with products
How can we solve the system of differential equations
$\frac{d f (t)}{d t} = - f (t) h (t), \frac{d g (t)}{d t} = - g (t) h (t), \frac{d h (t)}{d t} = 1 - (h (t))^{2}$
The system does not fall to standard ODE methods.

Isaiah Owens 2022-05-21

solving equations by the method of elimination
$\frac{a}{x} + \frac{b}{y} = \frac{a}{2} + \frac{b}{3} \dots (i)$
$x + 1 = y \dots (i i)$
We have to solve for x and y, only this time using the method of elimination.
From equation $(i i)$ , we get,
$\frac{1}{x + 1} = \frac{1}{y} \Rightarrow \frac{b}{x + 1} = \frac{b}{y} \dots (i i i)$
Subtracting $(i i i)$ from $(i)$ , we get,
$\frac{a}{x} + \frac{b}{y} - \frac{b}{y} = \frac{a}{2} + \frac{b}{3} - \frac{b}{x + 1}$
$\Rightarrow \frac{a}{x} = \frac{a}{2} + \frac{b}{3} - \frac{b}{x + 1}$
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

$M = 1 + \frac{a}{b}$
$S = a + b$
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

Find m to the equation:
${\begin{cases} 2 x^{3} - (y + 2) x^{2} + x y = m (1) \\ x^{2} + x - y = 1 - 2 m (2) \end{cases}$
My try: From $(1)$ and $(2) \Rightarrow$ :
$4 x^{3} - 2 (y + 2) x^{2} + 2 x y + x^{2} + x - y = 1 \Leftrightarrow 4 x^{3} - 3 x^{2} + x - 1 = y (2 x^{2} - 2 x + 1) \Leftrightarrow y = \frac{4 x^{3} - 3 x^{2} + x - 1}{2 x^{2} - 2 x + 1} \Leftrightarrow y = 2 x + \frac{1}{2} - \frac{3}{2 (2 x^{2} - 2 x + 1)}$
From $(2) \Rightarrow$
$2 m = 1 + y - x^{2} - x = 1 + 2 x + \frac{1}{2} - \frac{3}{2 (2 x^{2} - 2 x + 1)} - x^{2} - x = - x^{2} + x + \frac{3}{2} - \frac{3}{2 (2 x^{2} - 2 x + 1)}$
And I don't know how to contine,
The result is: $m \leq 1 - \frac{\sqrt{3}}{2}$

Shamar Reese 2022-05-21

Solve following series of equations ( $n + 2$ equations $n + 2$ variables):
$k_{0} q_{0} + λ q_{0} + c_{0} = 0, k_{1} q_{1} + λ q_{1} + c_{1} = 0, k_{n} q_{n} + λ q_{n} + c_{n} = 0, q_{1} + q_{2} + . . . . + q_{n} = 1.$
The variables are $q_{0}, q_{1}, . . . . ., q_{n}$ and $λ$ . Note that $k$ and $c$ are series of constants.

Trevor Wood 2022-05-21

Applications of polynomial systems of equations
What are some applications of Polynomial Systems of Equations?

Marianna Stone 2022-05-21

How to solve coupled linear 1st order PDE
It is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if
$\partial_{t} f + a \partial_{x} f = b f$
we have that $\frac{d f}{d t} = b f$ on the characteristic curve of $\frac{d x}{d t} = a$ . From this we deduce that $f (t, x) = g (C) e^{b t}$ where $x = a t + C$ .
Now, how does this work when $f$ is multidimensional. Can I solve equations on the following form by characteristics, or by any other means?
$\partial_{t} f_{i} (t, x) + \sum_{j} A_{i j} \partial_{x} f_{j} (t, x) = \sum_{j} B_{i j} f_{j} (t, x)$
where the components of $A$ and $B$ might be dependent on $x$ and $t$ .
In particular, I am trying to solve the following,
${\begin{cases} \partial_{t} f + \frac{c}{t} \partial_{x} g = - (a + \frac{1}{t}) f \\ \partial_{t} g + \frac{c}{t} \partial_{x} f = - (b + \frac{1}{t}) g \end{cases}$
where $f$ and $g$ are functions of $x$ and $t$ , where $t > t_{0} > 0$ , $c \neq 0$ . Any help is highly appreciated.

Davin Fields 2022-05-21

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

system of equations:
$\sqrt{x} + y = 7$
$\sqrt{y} + x = 11$
Its pretty visible that the solution is $(x, y) = (9, 4)$
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 $2 * 2$ , $4 * 1$ or $1 * 4$ . 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?

infogus88 2022-05-21

$x_{1} (t + 1) = (1 - m ρ_{1}) x_{1} (t) + n ρ_{2} x_{2} (t) + h 1$
$x_{2} (t + 1) = (1 - m ρ_{2}) x_{2} (t) + n ρ_{1} x_{1} (t) + h 2$
Suppose $x_{1} (0)$ and $x_{2} (0)$ are known. How can I find the analytical form of $x_{1} (t)$ and $x_{2} (t)$ ? Without the recursion it is
$x_{1} (t) = (1 - m ρ_{1})^{t} x_{1} (0) + \frac{1 - (1 - m ρ_{1})^{t}}{m ρ_{1}} h_{1}$
but the recursive form makes it too complicated.

skottyrottenmf 2022-05-21

System of nonlinear differential equations
${\dot{y}}_{1} = 2 - y_{1} y_{2} - y_{2}^{2}$
${\dot{y}}_{2} = 2 - y_{1}^{2} - y_{1} y_{2}$
i) Calculate the equilibrium points en determine their stability.
ii) Draw the Phase Plot.

Avah Knapp 2022-05-21

Test the consistency of the system of linear equations
$\begin{aligned} 4 x - 5 y + z & = 2 \\ 3 x + y - 2 z & = 9 \\ x + 4 y + z & = 5 \end{aligned}$

hoperetauyk 2022-05-21

find all the values of a and b so that the system has
a) no solution
b) 1 solution
c) exactly 3 solutions
d) infinitely many solutions
${\begin{cases} x & - & y & + & 2 z & = 4 \\ 3 x & - & 2 y & + & 9 z & = 14 \\ 2 x & - & 4 y & + & a z & = b \end{cases}$
I know that $a$ and $b$ has to either equal to something or not in order to satisfy the $4$ conditions stated above.
my matrices looked like
$[\begin{matrix} 1 & - 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & a + 12 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 6 \\ 2 \\ b + 8 \end{matrix}]$

Andy Erickson 2022-05-20

System of Differential Equations for Particular Initial Conditions
${\begin{cases} x^{'} (t) = x (t) + y (t) \\ y^{'} (t) = y (t) \end{cases}$
With the intial conditions $x (0) = 1$ and $y (0) = 1.$
Obviously, $y (t)$ (as well as $y^{'} (t)$ are equal to $e^{t}$ .
How do I find $x (t)$ now?
I have tried the following:
$x^{'} (t) = x (t) + y (t)$ can be written as
$y (t) = x^{'} (t) - x (t) .$
Now we know that $x (0)$ and $y (0) = 1$ , hence: $1 = x^{'} (0) - 1, ∴ x^{'} (0) = 2.$ .
How do I proceed from here?

Laylah Mora 2022-05-20

Solve $x y = 3$ and $4^{x^{2}} + 2^{y^{2}} = 72$

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.

Algebra I

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