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Cx-1x=10 la ecuación

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Solución numérica:

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Solución

Ha introducido [src] 
c*x - x = 10
cx−x=10c x - x = 10
Simplificar
Solución detallada
Tenemos una ecuación lineal:
c*x-1*x = 10

Dividamos ambos miembros de la ecuación en (-x + c*x)/x
x = 10 / ((-x + c*x)/x)

Obtenemos la respuesta: x = 10/(-1 + c)
Resolución de la ecuación paramétrica
Se da la ecuación con parámetro:
cx−x=10c x - x = 10
Коэффициент при x равен
c−1c - 1
entonces son posibles los casos para c :
c<1c < 1
c=1c = 1
Consideremos todos los casos con detalles:
Con
c<1c < 1
la ecuación será
−x−10=0- x - 10 = 0
su solución
x=−10x = -10
Con
c=1c = 1
la ecuación será
−10=0-10 = 0
su solución
no hay soluciones
Gráfica
Suma y producto de raíces [src] 
suma
   10*(-1 + re(c))             10*I*im(c)      
---------------------- - ----------------------
            2     2                  2     2   
(-1 + re(c))  + im (c)   (-1 + re(c))  + im (c)
10(re⁡(c)−1)(re⁡(c)−1)2+(im⁡(c))2−10iim⁡(c)(re⁡(c)−1)2+(im⁡(c))2\frac{10 \left(\operatorname{re}{\left(c\right)} - 1\right)}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} - \frac{10 i \operatorname{im}{\left(c\right)}}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} 
=
   10*(-1 + re(c))             10*I*im(c)      
---------------------- - ----------------------
            2     2                  2     2   
(-1 + re(c))  + im (c)   (-1 + re(c))  + im (c)
10(re⁡(c)−1)(re⁡(c)−1)2+(im⁡(c))2−10iim⁡(c)(re⁡(c)−1)2+(im⁡(c))2\frac{10 \left(\operatorname{re}{\left(c\right)} - 1\right)}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} - \frac{10 i \operatorname{im}{\left(c\right)}}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} 
producto
   10*(-1 + re(c))             10*I*im(c)      
---------------------- - ----------------------
            2     2                  2     2   
(-1 + re(c))  + im (c)   (-1 + re(c))  + im (c)
10(re⁡(c)−1)(re⁡(c)−1)2+(im⁡(c))2−10iim⁡(c)(re⁡(c)−1)2+(im⁡(c))2\frac{10 \left(\operatorname{re}{\left(c\right)} - 1\right)}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} - \frac{10 i \operatorname{im}{\left(c\right)}}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} 
=
10*(-1 - I*im(c) + re(c))
-------------------------
              2     2    
  (-1 + re(c))  + im (c)
10(re⁡(c)−iim⁡(c)−1)(re⁡(c)−1)2+(im⁡(c))2\frac{10 \left(\operatorname{re}{\left(c\right)} - i \operatorname{im}{\left(c\right)} - 1\right)}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} 
10*(-1 - i*im(c) + re(c))/((-1 + re(c))^2 + im(c)^2)
Respuesta rápida [src] 
        10*(-1 + re(c))             10*I*im(c)      
x1 = ---------------------- - ----------------------
                 2     2                  2     2   
     (-1 + re(c))  + im (c)   (-1 + re(c))  + im (c)
x1=10(re⁡(c)−1)(re⁡(c)−1)2+(im⁡(c))2−10iim⁡(c)(re⁡(c)−1)2+(im⁡(c))2x_{1} = \frac{10 \left(\operatorname{re}{\left(c\right)} - 1\right)}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} - \frac{10 i \operatorname{im}{\left(c\right)}}{\left(\operatorname{re}{\left(c\right)} - 1\right)^{2} + \left(\operatorname{im}{\left(c\right)}\right)^{2}} 
x1 = 10*(re(c) - 1)/((re(c) - 1)^2 + im(c)^2) - 10*i*im(c)/((re(c) - 1)^2 + im(c)^2)
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