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Resolver el siguiente sistema de Ecuaciones no Lineal:

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\$\\begin{eqnarray*}\\simplify{{y}}&=&\\simplify{x^2+{a}x+{a*b-c}}\\\\\\\\\\simplify{{y}}&=&\\simplify{-{b}x-{c}}\\end{eqnarray*}\$

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", "ungrouped_variables": ["a", "b", "c", "a1", "b1", "c1", "d1", "f1", "g1", "h1", "a2", "b2", "c2", "d2", "f2", "g2", "h2", "d"], "name": "Sistema de Ecuaciones no Lineal (3).jk", "functions": {}, "metadata": {"description": "

Shows how to define variables to stop degenerate examples.

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Observación: Para ingresar su respuesta como pares ordenados, hágalo con la siguiente condición: $\\,x_1<x_2$

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$x_1=$ [[0]] , $y_1=$ [[2]]

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y

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$x_2=$ [[1]] , $y_2=$ [[3]]

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"}], "extensions": [], "tags": [], "advice": "

Igualemos las ecuaciones y ordenemos:

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\\begin{align}
\\simplify{x^2+{a}x+{a*b-c}}=&\\,\\simplify{-{b}x-{c}}
\\end{align}

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\\begin{align}
\\simplify{x^2+{a+b}x+{a*b}}=&\\,\\simplify{0}
\\end{align}

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Factoricemos y resolvamos:

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\\begin{align}
(\\simplify{(x+{a})})(\\simplify{x+{b}})=&\\simplify{0}
\\end{align}

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\$\\simplify{(x+{a}={0})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{(x+{b}={0})} \$

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\$\\simplify{(x_1={-a})} \\,\\,\\,\\,;\\,\\,\\, \\simplify{(x_2={-b})} \$

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Ahora para $\\simplify{(x_1={-a})}$, debemos encontrar su respectivo valor de $\\,\\,y_1$. Para eso reemplazanos en la segunda ecuación del sistema inicial:

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\\begin{align}
\\simplify{y}=&\\,\\simplify{-{b}x-{c}} \\\\
\\simplify{y_1}=&\\,\\simplify[!collectnumbers]{-{b}*-{a}-{c}} \\\\
\\simplify{y_1}=&\\,\\simplify{-{b}{-a}-{c}}
\\end{align}

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Nuestra primera solución es el par ordenado:

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\\begin{align}
{(x_1,y_1)}=&\\,(\\simplify{{-a}},\\simplify{-{b}{-a}-{c}})
\\end{align}

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Para $\\simplify{(x_2={-b})}$, debemos encontrar su respectivo valor de $\\,\\,y_2$. Volvemos a reemplazar en la segunda ecuación del sistema inicial:

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\\begin{align}
\\simplify{y}=&\\,\\simplify{-{b}x-{c}} \\\\
\\simplify{y_2}=&\\,\\simplify[!collectnumbers]{-{b}*-{b}-{c}} \\\\
\\simplify{y_1}=&\\,\\simplify{-{b}{-b}-{c}}
\\end{align}

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La segunda solución es el par ordenado:

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\\begin{align}
{(x_2,y_2)}=&\\,(\\simplify{{-b}},\\simplify{-{b}{-b}-{c}})
\\end{align}

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Finalmente las soluciones del sistema son:

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\${(x_1,y_1)}=(\\simplify{{-a}},\\simplify{-{b}{-a}-{c}}) \\,\\,\\,\\, \\text{y}\\,\\,\\, {(x_2,y_2)}=(\\simplify{{-b}},\\simplify{-{b}{-b}-{c}}) \$

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