How to solve a system of two linear equations with two unknowns?

$$\{\begin{array}{l}7(a+b)=b-a\\ 4(3a+2b)=b-8\end{array}$$

$$\{\begin{array}{l}7(a+b)=b-a\\ 4(3a+2b)=b-8\end{array}$$

easternerjx
2022-09-26
Answered

How to solve a system of two linear equations with two unknowns?

$$\{\begin{array}{l}7(a+b)=b-a\\ 4(3a+2b)=b-8\end{array}$$

$$\{\begin{array}{l}7(a+b)=b-a\\ 4(3a+2b)=b-8\end{array}$$

You can still ask an expert for help

ruinsraidy4

Answered 2022-09-27
Author has **17** answers

Here is a different approach:

$\{\begin{array}{l}7(a+b)=b-a\\ 4(3a+2b)=b-8\end{array}$

Then

$\left[\begin{array}{ccc}8& 6& 0\\ 12& 7& -8\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 3/4& 0\\ 12& 7& -8\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 3/4& 0\\ 0& -2& -8\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 3/4& 0\\ 0& 1& 4\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\end{array}\right]$

Resulting in $a=-3$ and $b=4$.

$\{\begin{array}{l}7(a+b)=b-a\\ 4(3a+2b)=b-8\end{array}$

Then

$\left[\begin{array}{ccc}8& 6& 0\\ 12& 7& -8\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 3/4& 0\\ 12& 7& -8\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 3/4& 0\\ 0& -2& -8\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 3/4& 0\\ 0& 1& 4\end{array}\right]$ $\iff $ $\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\end{array}\right]$

Resulting in $a=-3$ and $b=4$.

Lustyku8

Answered 2022-09-28
Author has **2** answers

Try to perform algebra on both the equations till you have the 𝑎 and 𝑏 on one side and a number on the other. Then see if you can "combine" them together.

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 2022-06-23

Solving a linear system of equations

$\{\begin{array}{l}3x-2y+z=8\\ 4x-y+3z=-1\\ 5x+y+2z=-1\end{array}$

Form two equations with $y$ elimanted.

$\{\begin{array}{l}3x-2y+z=8\\ 4x-y+3z=-1\\ 5x+y+2z=-1\end{array}$

Form two equations with $y$ elimanted.

asked 2022-10-06

We are required to solve the following system of equations:

$$\begin{array}{}\text{(1)}& {x}^{3}+\frac{1}{3{x}^{4}}=5\end{array}$$

$$\begin{array}{}\text{(2)}& {x}^{4}+\frac{1}{3{x}^{3}}=10\end{array}$$

We may multiply (1) by $3{x}^{4}$ throughout and (2) by $3{x}^{3}$ throughout (as 0 is not a solution we may cancel the denominators) to yield.

$$\begin{array}{}\text{(3)}& 3{x}^{7}+1=15{x}^{4}\end{array}$$

$$\begin{array}{}\text{(4)}& 3{x}^{7}+1=30{x}^{3}\end{array}$$

Subtracting the two:

$$15{x}^{4}-30{x}^{3}=0$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}(x-2)=$$

As $0$ is not a solution we choose $x=2$.

But putting $x=2$ in the original equations does not satisfy them. How come?

$$\begin{array}{}\text{(1)}& {x}^{3}+\frac{1}{3{x}^{4}}=5\end{array}$$

$$\begin{array}{}\text{(2)}& {x}^{4}+\frac{1}{3{x}^{3}}=10\end{array}$$

We may multiply (1) by $3{x}^{4}$ throughout and (2) by $3{x}^{3}$ throughout (as 0 is not a solution we may cancel the denominators) to yield.

$$\begin{array}{}\text{(3)}& 3{x}^{7}+1=15{x}^{4}\end{array}$$

$$\begin{array}{}\text{(4)}& 3{x}^{7}+1=30{x}^{3}\end{array}$$

Subtracting the two:

$$15{x}^{4}-30{x}^{3}=0$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}(x-2)=$$

As $0$ is not a solution we choose $x=2$.

But putting $x=2$ in the original equations does not satisfy them. How come?

asked 2021-12-13

What is the value of $x-y$ , if $xy=144$ , $x+y=30$ , and $x>y$ ?

asked 2022-05-14

Solving systems of equations in 3 variables

$x+y+z=0$

$xy+yz+xz=c-{c}^{2}$

$xyz=3{c}^{2}$

where, $c$ is some fixed non-zero constant.

$x+y+z=0$

$xy+yz+xz=c-{c}^{2}$

$xyz=3{c}^{2}$

where, $c$ is some fixed non-zero constant.

asked 2022-05-13

How to solve this simultaneous equation of 3 variables.

$\begin{array}{}\text{(1)}& x+y+z& =& a+b+c\text{(2)}& ax+by+cz& =& {a}^{2}+{b}^{2}+{c}^{2}\text{(3)}& a{x}^{2}+b{y}^{2}+c{z}^{2}& =& {a}^{3}+{b}^{3}+{c}^{3}\end{array}$

$\begin{array}{}\text{(1)}& x+y+z& =& a+b+c\text{(2)}& ax+by+cz& =& {a}^{2}+{b}^{2}+{c}^{2}\text{(3)}& a{x}^{2}+b{y}^{2}+c{z}^{2}& =& {a}^{3}+{b}^{3}+{c}^{3}\end{array}$

asked 2022-06-01

How do I find the equilibria of this autonomous system of ODEs?

${x}^{\prime}(t)=x({x}^{3}-2{y}^{3})$

${y}^{\prime}(t)=y(2{x}^{3}-{y}^{3})$

${x}^{\prime}(t)=x({x}^{3}-2{y}^{3})$

${y}^{\prime}(t)=y(2{x}^{3}-{y}^{3})$