Find answers of this system of equations in reals

$$\{\begin{array}{c}x+3y=4{y}^{3}\\ y+3z=4{z}^{3}\\ z+3x=4{x}^{3}\end{array}$$

$$\{\begin{array}{c}x+3y=4{y}^{3}\\ y+3z=4{z}^{3}\\ z+3x=4{x}^{3}\end{array}$$

David Ali
2022-09-20
Answered

Find answers of this system of equations in reals

$$\{\begin{array}{c}x+3y=4{y}^{3}\\ y+3z=4{z}^{3}\\ z+3x=4{x}^{3}\end{array}$$

$$\{\begin{array}{c}x+3y=4{y}^{3}\\ y+3z=4{z}^{3}\\ z+3x=4{x}^{3}\end{array}$$

You can still ask an expert for help

Nathalie Rivers

Answered 2022-09-21
Author has **7** answers

Suppose that we had $x>1$. Then, since $4{x}^{3}-3x>x$, we have $z>x$, and similarly $y>z$, $x>y$, contradiction. By symmetry, we conclude that $x,y,z\in [-1,1]$.

So there exist $\alpha ,\beta ,\gamma \in [0,\pi ]$ with $x=\mathrm{cos}\alpha $, $y=\mathrm{cos}\beta $, $z=\mathrm{cos}\gamma $. By the formula for $\mathrm{cos}3\alpha $, we can rewrite the system of equations as:

$$\{\begin{array}{c}\alpha \equiv \pm 3\beta \text{}(\mathrm{mod}\text{}2\pi )\\ \beta =\pm 3\gamma \text{}(\mathrm{mod}\text{}2\pi )\\ \gamma =\pm 3\alpha \text{}(\mathrm{mod}\text{}2\pi )\end{array}$$

So we have $\pm 27\alpha \equiv \alpha $, so either $26\alpha \equiv 0$ or $28\alpha \equiv 0$. We find that $\alpha =\pi k/13$ or $\alpha =\pi k/14$ for some nonnegative integer $k$.

This gives $27$ solutions, $x=\mathrm{cos}\frac{\pi k}{13}$ for $0\le k\le 13$, and $x=\mathrm{cos}\frac{\pi k}{14}$ for $1\le k\le 13$.

For example, one solution is ($(\mathrm{cos}\frac{\pi}{14},\mathrm{cos}\frac{9\pi}{14},\mathrm{cos}\frac{3\pi}{14})$).

So there exist $\alpha ,\beta ,\gamma \in [0,\pi ]$ with $x=\mathrm{cos}\alpha $, $y=\mathrm{cos}\beta $, $z=\mathrm{cos}\gamma $. By the formula for $\mathrm{cos}3\alpha $, we can rewrite the system of equations as:

$$\{\begin{array}{c}\alpha \equiv \pm 3\beta \text{}(\mathrm{mod}\text{}2\pi )\\ \beta =\pm 3\gamma \text{}(\mathrm{mod}\text{}2\pi )\\ \gamma =\pm 3\alpha \text{}(\mathrm{mod}\text{}2\pi )\end{array}$$

So we have $\pm 27\alpha \equiv \alpha $, so either $26\alpha \equiv 0$ or $28\alpha \equiv 0$. We find that $\alpha =\pi k/13$ or $\alpha =\pi k/14$ for some nonnegative integer $k$.

This gives $27$ solutions, $x=\mathrm{cos}\frac{\pi k}{13}$ for $0\le k\le 13$, and $x=\mathrm{cos}\frac{\pi k}{14}$ for $1\le k\le 13$.

For example, one solution is ($(\mathrm{cos}\frac{\pi}{14},\mathrm{cos}\frac{9\pi}{14},\mathrm{cos}\frac{3\pi}{14})$).

asked 2020-11-08

2x + 3y = 34

y = 5x

y = 5x

asked 2022-07-11

Sum of cubed roots

${x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}$

${x}_{1}^{4}+{x}_{2}^{4}+{x}_{3}^{4}$

where ${x}_{1},{x}_{2},{x}_{3}$ are the roots of

${x}^{3}+2{x}^{2}+3x+4=0$

using Viete's formulas.

I know that ${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=-2$, as I already calculated that, but I can't seem to get the cube of the roots. I've tried

$({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2})({x}_{1}+{x}_{2}+{x}_{3})$

but that did work.

${x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}$

${x}_{1}^{4}+{x}_{2}^{4}+{x}_{3}^{4}$

where ${x}_{1},{x}_{2},{x}_{3}$ are the roots of

${x}^{3}+2{x}^{2}+3x+4=0$

using Viete's formulas.

I know that ${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=-2$, as I already calculated that, but I can't seem to get the cube of the roots. I've tried

$({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2})({x}_{1}+{x}_{2}+{x}_{3})$

but that did work.

asked 2022-05-23

For which value(s) of parameter m is there a solution for this system

$\{\begin{array}{l}mx+y=m\\ mx+2y=1\\ 2x+my=m+1\end{array}$

when does this system of equations have a solution?

$\{\begin{array}{l}mx+y=m\\ mx+2y=1\\ 2x+my=m+1\end{array}$

when does this system of equations have a solution?

asked 2022-10-27

The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, the denominator becomes 12 times the numerator. Determine the fraction.

Let the numerator be $$\mathit{x}$$ and the denominator be $$\mathit{y}$$

Therefore, Fraction $$=\frac{x}{4}+2x$$

Because, both the numerator and denominator are decreased by $$6$$

Therefore, the new fraction becomes $$\mathit{x}-\frac{6}{(4+2\mathit{x})}-6=12\mathit{x}$$

Let the numerator be $$\mathit{x}$$ and the denominator be $$\mathit{y}$$

Therefore, Fraction $$=\frac{x}{4}+2x$$

Because, both the numerator and denominator are decreased by $$6$$

Therefore, the new fraction becomes $$\mathit{x}-\frac{6}{(4+2\mathit{x})}-6=12\mathit{x}$$

asked 2021-06-23

Write

a.

b.

c.

asked 2022-10-08

Renata walks down an escalator that moves up and counts 150 steps. Her sister Fernanda climbs the same escalator and counts 75 steps. If the speed of Renata (in steps per time unit) is three times the speed of Fernanda, determine how many steps are visible on the escalator at any time.

asked 2022-11-18

Solving system of equations: ${x}^{3}-3{y}^{2}x=-1$ and $3y{x}^{2}-{y}^{3}=1$