Solve the system of equations in response write down the sum of the solutions of the system {4x-2y=2 and below 2x+y=5

tamolam8
2022-09-12
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asked 2022-06-21

Find all complex values for $a$ where there's no solution for the non homogeneous system

$\{\begin{array}{c}x+(1-a)y+z=1\\ 2x-y+z=3\\ 3x-ay+(a-1)z=4\end{array}$

After moving to a matrix representation and reduction, I have:

$\left(\begin{array}{ccccc}1& 1-a& 1& |& 1\\ 0& 2a-3& -1& |& 1\\ 0& 0& a-1& |& 0\end{array}\right)$

Now, I don't see an $a$ that answering this request, and I double-checked my reduction process.

$\{\begin{array}{c}x+(1-a)y+z=1\\ 2x-y+z=3\\ 3x-ay+(a-1)z=4\end{array}$

After moving to a matrix representation and reduction, I have:

$\left(\begin{array}{ccccc}1& 1-a& 1& |& 1\\ 0& 2a-3& -1& |& 1\\ 0& 0& a-1& |& 0\end{array}\right)$

Now, I don't see an $a$ that answering this request, and I double-checked my reduction process.

asked 2022-07-02

Systems of equations question

$\begin{array}{rl}{a}^{a}\cdot {b}^{b}\cdot {c}^{c}\cdot {d}^{d}& =\frac{1}{2}\\ a+b+c+d& =1\end{array}$

How can we find solutions for this system of equations given that $a,b,c,d>0$?

$\begin{array}{rl}{a}^{a}\cdot {b}^{b}\cdot {c}^{c}\cdot {d}^{d}& =\frac{1}{2}\\ a+b+c+d& =1\end{array}$

How can we find solutions for this system of equations given that $a,b,c,d>0$?

asked 2021-09-14

Solve the following system of equations:

asked 2022-06-14

Have the following system of equations:

$2+{x}^{2}-{y}^{2}=0$

${x}^{2}-{y}^{2}-2=0$

And if I substitute y by a function of x and vice versa I get:

$2+{x}^{2}-{x}^{2}+2=0$

${y}^{2}-{y}^{2}-4=0$

I therefore get:

$4=0,-4=0$ Therefore I don't have any solutions for that system of equations

In theory, am I allowed to get to this conclusion?

$2+{x}^{2}-{y}^{2}=0$

${x}^{2}-{y}^{2}-2=0$

And if I substitute y by a function of x and vice versa I get:

$2+{x}^{2}-{x}^{2}+2=0$

${y}^{2}-{y}^{2}-4=0$

I therefore get:

$4=0,-4=0$ Therefore I don't have any solutions for that system of equations

In theory, am I allowed to get to this conclusion?

asked 2022-05-21

${x}_{1}(t+1)=(1-m{\rho}_{1}){x}_{1}(t)+n{\rho}_{2}{x}_{2}(t)+h1$

${x}_{2}(t+1)=(1-m{\rho}_{2}){x}_{2}(t)+n{\rho}_{1}{x}_{1}(t)+h2$

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

${x}_{1}(t)=(1-m{\rho}_{1}{)}^{t}{x}_{1}(0)+\frac{1-(1-m{\rho}_{1}{)}^{t}}{m{\rho}_{1}}{h}_{1}$

but the recursive form makes it too complicated.

${x}_{2}(t+1)=(1-m{\rho}_{2}){x}_{2}(t)+n{\rho}_{1}{x}_{1}(t)+h2$

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

${x}_{1}(t)=(1-m{\rho}_{1}{)}^{t}{x}_{1}(0)+\frac{1-(1-m{\rho}_{1}{)}^{t}}{m{\rho}_{1}}{h}_{1}$

but the recursive form makes it too complicated.

asked 2022-06-01

How to solve this sytem of equation:

$\{\begin{array}{l}(\sqrt{{a}^{2}+4}+a)(\sqrt{{b}^{2}+1}+b)=1\\ 27{a}^{6}+8b={a}^{3}+2\end{array}$

I tried to analysis equation (1):

$\sqrt{{a}^{2}+4}+a=\sqrt{{b}^{2}+1}-b\iff a+b=\sqrt{{b}^{2}+1}-\sqrt{{a}^{2}+4}$

$\iff 4{a}^{2}+16{b}^{2}+20ab=9$

And then, I don't know how to solve this.

$\{\begin{array}{l}(\sqrt{{a}^{2}+4}+a)(\sqrt{{b}^{2}+1}+b)=1\\ 27{a}^{6}+8b={a}^{3}+2\end{array}$

I tried to analysis equation (1):

$\sqrt{{a}^{2}+4}+a=\sqrt{{b}^{2}+1}-b\iff a+b=\sqrt{{b}^{2}+1}-\sqrt{{a}^{2}+4}$

$\iff 4{a}^{2}+16{b}^{2}+20ab=9$

And then, I don't know how to solve this.

asked 2022-05-21

Test the consistency of the system of linear equations

$\begin{array}{rl}4x-5y+z& =2\\ 3x+y-2z& =9\\ x+4y+z& =5\end{array}$

$\begin{array}{rl}4x-5y+z& =2\\ 3x+y-2z& =9\\ x+4y+z& =5\end{array}$