Solve the system:

Tazmin Horton
2021-02-10
Answered

Solve the system:

You can still ask an expert for help

cheekabooy

Answered 2021-02-11
Author has **83** answers

Multiply equation (1) by 3:

The lines (1) and (2) are parallel, hence the system doesn’t have a solution.

asked 2022-05-27

Solve a system of two linear equations

If $2x+3y=\frac{1}{10}$ and $3x+2y=\frac{1}{8}$, then how to calculate here $x$ and $y$?

If $2x+3y=\frac{1}{10}$ and $3x+2y=\frac{1}{8}$, then how to calculate here $x$ and $y$?

asked 2022-05-30

Is there any nonempty, compact and invariant set in dynamical system generated by this system of equations?

${x}^{\prime}=x+\mathrm{sin}(xy+2)-7$

My idea is to use this fact: Not empty omega limit set - because here we have also bounded functions and omega limit set is invariant. But it's hard to say anything about compactness.

${x}^{\prime}=x+\mathrm{sin}(xy+2)-7$

My idea is to use this fact: Not empty omega limit set - because here we have also bounded functions and omega limit set is invariant. But it's hard to say anything about compactness.

asked 2022-11-24

Let $a,b,c$ be nonzero real numbers and let ${a}^{2}-{b}^{2}=bc$ and ${b}^{2}-{c}^{2}=ca$. Prove that ${a}^{2}-{c}^{2}=ab$.

The solution strategy given in the course was to scale the two given equations by $s=\frac{1}{c}$, resulting in ${a}^{2}-{b}^{2}=bc$ becoming ${a}^{2}-{b}^{2}=b$ and ${b}^{2}-{c}^{2}=ca$ becoming ${b}^{2}-1=a$. $c$ is basically being set to 1, but I don't understand the justification. Doesn't scaling by $\frac{1}{c}$ by definition not change the equations, since $(a/c{)}^{2}-(b/c{)}^{2}=(b/c)(c/c)\u27fa{\displaystyle \frac{{a}^{2}-{b}^{2}}{{c}^{2}}}={\displaystyle \frac{bc}{{c}^{2}}}\u27fa{a}^{2}-{b}^{2}=bc$

Where is mistake?

The solution strategy given in the course was to scale the two given equations by $s=\frac{1}{c}$, resulting in ${a}^{2}-{b}^{2}=bc$ becoming ${a}^{2}-{b}^{2}=b$ and ${b}^{2}-{c}^{2}=ca$ becoming ${b}^{2}-1=a$. $c$ is basically being set to 1, but I don't understand the justification. Doesn't scaling by $\frac{1}{c}$ by definition not change the equations, since $(a/c{)}^{2}-(b/c{)}^{2}=(b/c)(c/c)\u27fa{\displaystyle \frac{{a}^{2}-{b}^{2}}{{c}^{2}}}={\displaystyle \frac{bc}{{c}^{2}}}\u27fa{a}^{2}-{b}^{2}=bc$

Where is mistake?

asked 2022-10-06

If ${x}^{a}+{x}^{b}={x}^{c}+{x}^{d}$ how do $a,b,c,d$ relationship are?

For instance:

$${3}^{(3{x}^{2}+8)}+{3}^{(4x+2)}={3}^{0}+{3}^{(5{x}^{2}+7)}$$

and

$$(3{x}^{2}+8)(4x+2)\ne 0$$

For instance:

$${3}^{(3{x}^{2}+8)}+{3}^{(4x+2)}={3}^{0}+{3}^{(5{x}^{2}+7)}$$

and

$$(3{x}^{2}+8)(4x+2)\ne 0$$

asked 2022-06-14

Solving 3 simultaneous cubic equations

${i}_{1}^{3}{L}_{1}+{i}_{1}K+{V}_{1}+({i}_{2}+{i}_{3}+C){Z}_{n}=0$

${i}_{2}^{3}{L}_{2}+{i}_{2}K+{V}_{2}+({i}_{1}+{i}_{3}+C){Z}_{n}=0$

${i}_{3}^{3}{L}_{3}+{i}_{3}K+{V}_{3}+({i}_{1}+{i}_{2}+C){Z}_{n}=0$

where ${L}_{1},{L}_{2},{L}_{3},K,{V}_{1},{V}_{2},{V}_{3},C$ and ${Z}_{n}$ are all known constants.

What methods can I use to obtain the values of ${i}_{1},{i}_{2}$ and${i}_{3}$?

${i}_{1}^{3}{L}_{1}+{i}_{1}K+{V}_{1}+({i}_{2}+{i}_{3}+C){Z}_{n}=0$

${i}_{2}^{3}{L}_{2}+{i}_{2}K+{V}_{2}+({i}_{1}+{i}_{3}+C){Z}_{n}=0$

${i}_{3}^{3}{L}_{3}+{i}_{3}K+{V}_{3}+({i}_{1}+{i}_{2}+C){Z}_{n}=0$

where ${L}_{1},{L}_{2},{L}_{3},K,{V}_{1},{V}_{2},{V}_{3},C$ and ${Z}_{n}$ are all known constants.

What methods can I use to obtain the values of ${i}_{1},{i}_{2}$ and${i}_{3}$?

asked 2022-10-23

If $x+y+z=12$ and ${x}^{2}+{y}^{2}+{z}^{2}=54$ then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.

asked 2022-06-26

Condition number of system of non-linear equations. The system has only two unknowns but 6 equations (thus over-determined). Solving the system of equations are not a problem. However, I need an indication of how well-conditioned the system of equations is. I know the condition number is typically used to do this. Any advice on exactly how this procedure works will be appreciated.