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I was recently doing some calculations and I have an equation pair (system of equations) which I wouuld like to be able solve (preferrably manually?)

Can this system of EQs be solved easily with pen-and-paper /AND/ OR basic calculator? (not a super-duper-calculator)

My old calculator cannot handle 2nd degree simulatenous equations for example.

problem system of equations

  • eq1 : $0 + 272.91 + \frac{800x}{ \sqrt{(x^2+y^2+900)}} = 0$
  • eq2 : $ -471.5 + 682.22 + \frac{800y}{\sqrt{(x^2+y^2+900)}} = 0$

I already solved this equation using my highly advanced TI Nspire CX CAS calculator but I was a little bit disappointed at not being able to solve it with pen-and-paper.

I have some confidence in Wolfram Alpha's solutions though, because I was able to double check my earlier work done in pen-and-paper using the Wolfram Alpha's solution. My earlier work was accurate up to 2 decimal places which ought to be ok.

If you really want to know how I came up with the system of Equations.

  • eq1 represents (three coefficients of i-basevector summed together) = 0

  • eq2 represents (three coefficients of j-basevector summed together) = 0

According to Wolfram Alpha the results are x= -11.3416 y= -8.75708

using those x,y and some other background information, I was able to formulate the third vector exactly. Then I calculated a vector sum with three vectors, and analyzed the vector sum by looking at the coefficients of the basevectors.

EDIT YES THERE WAS TYPO MINUS SIGN MISSING sorry it is supposed to be for

  • eq2 : $ -471.5 + 682.22 + \frac{800y}{\sqrt{(x^2+y^2+900)}} = 0$
  • eq2 : $ \leftrightarrow +210.72 + \frac{800y}{\sqrt{(x^2+y^2+900)}} = 0$
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  • $\begingroup$ (1) Let $u=x^2+y^2$ (2) Solve for $u$ in each equation. (3) Use the resulting expressions to solve for $x$ and $y$. $\endgroup$ – Paul Apr 17 '17 at 16:00
  • $\begingroup$ There is a strong preference on this site to use MathJax: math.meta.stackexchange.com/questions/5020/…. Second it can be solved, but it depends on what you mean by "easily" $\endgroup$ – Χpẘ Apr 17 '17 at 16:01
  • $\begingroup$ i have found only complex solutions $\endgroup$ – Dr. Sonnhard Graubner Apr 17 '17 at 16:03
  • $\begingroup$ is there a typo? $\endgroup$ – Dr. Sonnhard Graubner Apr 17 '17 at 16:03
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Hint:

Switching to polar coordinates, you can write

$$\frac{r\cos\theta}{\sqrt{r^2+900}}=a,\\\frac{r\sin\theta}{\sqrt{r^2+900}}=b.$$

Then

$$\tan\theta=\frac ba$$

and

$$\frac{r^2}{r^2+900}=a^2+b^2$$ leads to the solution.

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