0
$\begingroup$

$$\sin(x+y)+1.6x=0$$

$$x^2+y^2=-1$$

Can this system be solved? Please help me with it. I managed to make graphs of it but can't get it solved without graph.

Graph: enter image description here

$\endgroup$
  • $\begingroup$ in which domain is that to be solve? $\endgroup$ – Dr. Sonnhard Graubner Nov 22 '14 at 17:52
  • $\begingroup$ I think no imaginary numbers only. I will need to use Newton method after solution to find intersection. $\endgroup$ – Neone Nov 22 '14 at 17:54
3
$\begingroup$

Hint: For what real $x$ and $y$ will we have $$x^2+y^2=-1,$$ if any?

$\endgroup$
0
$\begingroup$

If the domain is real number no square numbers sum will be negative.

So no x and y satisfy the equation

$\endgroup$
0
$\begingroup$

since we have for all real numbers $x^2+y^2\geq 0>-1$ we get no real solutions for this system. you can only plug $y=\sqrt{-x^2-1}$ or $y=-\sqrt{-x^2-1}$ in your equation.

$\endgroup$
  • $\begingroup$ So it is not possible to use Newton method with this system? Or can we choose any x and y to start with? Because I have a graph. $\endgroup$ – Neone Nov 22 '14 at 17:59
  • $\begingroup$ it is also possible to use the newton method for complex numbers, or is a typo in your system? $\endgroup$ – Dr. Sonnhard Graubner Nov 22 '14 at 18:01
  • $\begingroup$ And what would be solution if we will use complex? $\endgroup$ – Neone Nov 22 '14 at 18:02
0
$\begingroup$

Solved:
$$sin(x+y)+1.6x=0$$ $$x^2+y^2=-1$$
$$y=√(|-x^2-1|)$$

$$ sin(x+√(|-x^2-1|)+1.6*x=0$$

x=-0.393536

$$ y=√(|0.393536^2-1|)$$

y=0.91931

Looking at the graph it seems these are correct values. Thank you, especially Dr. Sonnhard Graubner.

$\endgroup$
  • $\begingroup$ Your graph is incorrect. The values of $x$ and $y$ that you found are approximate solutions to $x^2+y^2=1,$ not to $x^2+y^2=-1.$ From $y^2=-x^2-1,$ we can conclude that $|y|=\sqrt{-x^2-1},$ not that $y=\sqrt{|-x^2-1|}.$ $\endgroup$ – Cameron Buie Nov 22 '14 at 19:07
  • $\begingroup$ Approximate values is what I need because later I will find them as accurate as needed according to error value with Newton method. If there is no x and y values which correspondent to $$x^2+y^2=−1$$ then I will calculate with 1 instead because work has to be done anyway.. Maybe this was a mistake made when filling equations. $\endgroup$ – Neone Nov 22 '14 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.