# Solve system of equations

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

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

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:

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

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

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

So no x and y satisfy the equation

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.

• 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. – Neone Nov 22 '14 at 17:59
• it is also possible to use the newton method for complex numbers, or is a typo in your system? – Dr. Sonnhard Graubner Nov 22 '14 at 18:01
• And what would be solution if we will use complex? – Neone Nov 22 '14 at 18:02

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.

• 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 $y=\sqrt{|-x^2-1|}.$ – Cameron Buie Nov 22 '14 at 19:07
• 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. – Neone Nov 22 '14 at 19:33