Sr Examen

absolute(x+2)+absolute(y+1)=0 la ecuación

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Solución numérica:

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Solución

Ha introducido [src]
|x + 2| + |y + 1| = 0
$$\left|{x + 2}\right| + \left|{y + 1}\right| = 0$$
Gráfica
Respuesta rápida [src]
         //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
x1 = I*im|<                              | + re|<                              |
         \\    nan          otherwise    /     \\    nan          otherwise    /
$$x_{1} = \operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
x1 = re(Piecewise((-Abs(y + 1 - 2, |y + 1| <= 0), (nan, True))) + i*im(Piecewise((-|y + 1| - 2, |y + 1| <= 0), (nan, True))))
Suma y producto de raíces [src]
suma
    //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
I*im|<                              | + re|<                              |
    \\    nan          otherwise    /     \\    nan          otherwise    /
$$\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
=
    //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
I*im|<                              | + re|<                              |
    \\    nan          otherwise    /     \\    nan          otherwise    /
$$\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
producto
    //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
I*im|<                              | + re|<                              |
    \\    nan          otherwise    /     \\    nan          otherwise    /
$$\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)}$$
=
/-(2 + |1 + y|)  for |1 + y| <= 0
<                                
\     nan           otherwise    
$$\begin{cases} - (\left|{y + 1}\right| + 2) & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}$$
Piecewise((-(2 + |1 + y|), |1 + y| <= 0), (nan, True))