Un sistema de ecuaciones es no lineal cuándo al menos una de sus ecuaciones no es de primer grado.
Ejemplo:
![Rendered by QuickLaTeX.com \left\{\begin{matrix} x^2+y^2=25\\ x+y=7 \ \ \ \ \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-8324fd36948ecf0832a6b2271296f1b6_l3.png)
PASOS DEL MÉTODO DE SUSTITUCIÓN
La resolución de estos sistemas se suele hacer por el método de sustitución, para ello seguiremos los siguientes pasos:
![Rendered by QuickLaTeX.com \left\{\begin{matrix} x^2+y^2=25\\ x+y=7 \ \ \ \ \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-8324fd36948ecf0832a6b2271296f1b6_l3.png)
1º Se despeja una incógnita en una de las ecuaciones, preferentemente en la de primer grado.
![Rendered by QuickLaTeX.com y=7-x](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-e862ab5ed3f3a55bff0fd7808907e06d_l3.png)
2º Se sustituye el valor de la incógnita despejada en la otra ecuación.
![Rendered by QuickLaTeX.com x^2+(7-x)^2=25](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-f81635939049906417aec2d1c4adda1f_l3.png)
3º Se resuelve la ecuación resultante.
![Rendered by QuickLaTeX.com x^2+49-14x+x^2=25](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-49e66261b5a0a3b9c20671ef1cc9d160_l3.png)
![Rendered by QuickLaTeX.com 2x^2-14x+24=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-bff976dea4cd92c207e782f63f4a8c63_l3.png)
![Rendered by QuickLaTeX.com x^2-7x+12=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-bb3f8c2c38a533d3002cc12b02c49950_l3.png)
![Rendered by QuickLaTeX.com \displaystyle x=\frac{7\pm \sqrt{49-48}}{2}=\frac{7\pm 1}{2}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-e9ab74334a95962b0c0348d46c89ad5c_l3.png)
![Rendered by QuickLaTeX.com x_1 =4](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-3181f890dc070cc74e3fb6388869fea0_l3.png)
![Rendered by QuickLaTeX.com x_2 =3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-16ed0a5d82ad7355a97adb1d4f06a855_l3.png)
4º Cada uno de los valores obtenidos se sustituyen en la otra ecuación. Se obtienen así los valores correspondientes de la otra incógnita.
![Rendered by QuickLaTeX.com x=3 \hspace{2cm} y=7-3 \hspace{2cm} y=4](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-302d6eaa5287546ff295a88480435c35_l3.png)
![Rendered by QuickLaTeX.com x=4 \hspace{2cm} y=7-4 \hspace{2cm} y=3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-9b35644bfd8c7cfa9106205b9d32caa4_l3.png)
EJERCICIOS RESUELTOS DE SISTEMAS DE ECUACIONES
![\left\{\begin{matrix} x+y=7\\ x\cdot y =12 \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-8f2b60ba67d562cf74521ef0843226ab_l3.png)
Solución1Despejamos una incógnita de la ecuación de primer grado
![Rendered by QuickLaTeX.com y=7-x](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-e862ab5ed3f3a55bff0fd7808907e06d_l3.png)
2 Sustituimos en la otra ecuación
![Rendered by QuickLaTeX.com x\cdot (7-x)=12](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-90a92d0b6e2ffa546497c2ba19991142_l3.png)
Desarrollamos
![Rendered by QuickLaTeX.com 7x-x^2=12](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-04230a5a0a4ccd7141a1f55d8e5ff5d8_l3.png)
Notamos que se trata de la ecuación cuadrática
![Rendered by QuickLaTeX.com x^2-7x+2=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-1adb5a7f0bc2cfe4ad284d4eec517c33_l3.png)
3 Resolvemos
Por fórmula general sabemos que
![Rendered by QuickLaTeX.com \displaystyle x=\frac{7\pm \sqrt{49-48}}{2}=\frac{7\pm 1}{2}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-cc7264896dab29a7d9afca095acefae2_l3.png)
![Rendered by QuickLaTeX.com x_1 =4](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-3181f890dc070cc74e3fb6388869fea0_l3.png)
![Rendered by QuickLaTeX.com x_2 =3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-16ed0a5d82ad7355a97adb1d4f06a855_l3.png)
4 Obtenemos el valor de la otra incógnita
![Rendered by QuickLaTeX.com x=4 \hspace{2cm} y=7-4 \hspace{2cm} y=3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-9b35644bfd8c7cfa9106205b9d32caa4_l3.png)
![Rendered by QuickLaTeX.com x=3 \hspace{2cm} y=7-3 \hspace{2cm} y=4](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-302d6eaa5287546ff295a88480435c35_l3.png)
————————————————
![\left\{\begin{matrix} x^2+y^2=169\\ x+y=17 \ \ \ \ \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-0ffea08cc78b407df5a167443e2f444a_l3.png)
Solución1Despejamos una incógnita de la ecuación de primer grado
![Rendered by QuickLaTeX.com x=17-y](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-5674c80cd9c4a3b7a32cd84ee93ba106_l3.png)
2 Sustituimos en la otra ecuación
![Rendered by QuickLaTeX.com (17-y)^2+y^2=169](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-f4a4dd18843b77ce53de724091cbc04f_l3.png)
Desarrollamos
![Rendered by QuickLaTeX.com 289-34y+y^2+y^2=169](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-1ea7e9a63d6266106186a6836cad3512_l3.png)
![Rendered by QuickLaTeX.com 2y^2 -34y+120=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-d29e8e8481c601a309801ea848cb10f1_l3.png)
Notamos que se trata de la ecuación cuadrática
![Rendered by QuickLaTeX.com y^2-17y+60=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-f8a31a8fe3d02c86aff7d9ed53280ea4_l3.png)
3 Resolvemos
Por fórmula general sabemos que
![Rendered by QuickLaTeX.com \displaystyle y=\frac{17\pm \sqrt{289-240}}{2}=\frac{17\pm 7}{2}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-15abe924643f34905dd61eb78bb9eb7b_l3.png)
![Rendered by QuickLaTeX.com y_1 =12](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-7bc9fcc90252ae7dcb140670aecf4ccf_l3.png)
![Rendered by QuickLaTeX.com y_2 =5](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-0fd68f908a37f32f35655b7c426dc580_l3.png)
4 Obtenemos el valor de la otra incógnita
![Rendered by QuickLaTeX.com y=12 \hspace{2cm} x=17-12 \hspace{2cm} x=5](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-ca76aa0f07ba4d37ecbc74aa9fe4e7ab_l3.png)
![Rendered by QuickLaTeX.com y=5 \hspace{2cm} x=17-5 \hspace{2cm} x=12](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-b3c6c5d6452cd4858a17cfa4acd985d1_l3.png)
————————————————
![\left\{\begin{matrix} y^2-2y+1=x\\ \sqrt{x}+y=5 \ \ \ \ \ \ \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-89866210abb82b0ab333b636ca728e90_l3.png)
Solución1Despejamos una incógnita de una de las ecuaciones
En este caso no hay de primer grado, pero notamos que x ya está despejada en la primera ecuación.
![Rendered by QuickLaTeX.com x=y^2-2y+1](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-b8e36b0e1b03f92a42661d777845f91a_l3.png)
2 Sustituimos en la otra ecuación
![Rendered by QuickLaTeX.com \sqrt{y^2-2y+1}+y=5](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-eb40f0e0c8460292335d820a9ac293ba_l3.png)
Despejamos la raíz
![Rendered by QuickLaTeX.com \sqrt{y^2-2y+1}=5-y](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-3609d63c09de583300b0a3830584cf82_l3.png)
Elevamos al cuadrado
![Rendered by QuickLaTeX.com \left(\sqrt{y^2-2y+1}\right)^2=(5-y)^2](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-329f62d612a5b119e060191694bb4be8_l3.png)
Desarrollamos
![Rendered by QuickLaTeX.com y^2-2y+1=25-10y+y^2](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-007a27b60bbe8c75a58c2fa188f8e789_l3.png)
![Rendered by QuickLaTeX.com 10y-2y=25-1](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-ed96330b1896973fc989190ce983c11a_l3.png)
3 Resolvemos
![Rendered by QuickLaTeX.com 8y=24](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-63244a0dd039d8045427e9fca479a259_l3.png)
![Rendered by QuickLaTeX.com y =3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-e33f04d4d239f724c9ca598a027dffa8_l3.png)
4 Obtenemos el valor de la otra incógnita
![Rendered by QuickLaTeX.com y=3 \hspace{2cm} x=3^2-2\cdot 3+1 \hspace{2cm} x=9-6+1=4](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-0224c0004915c99ea52a81f015986349_l3.png)
—————————————————
![\displaystyle \left\{\begin{matrix} \frac{1}{x^2}+\frac{1}{y^2}=13 \\ \\ \frac{1}{x}-\frac{1}{y}=1 \ \ \ \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-72c1689ee3f5396e1a84d27734d70073_l3.png)
Solución1Cambio de variableNotamos que bajo el cambio de variable
![Rendered by QuickLaTeX.com \displaystyle u=\frac{1}{x} \hspace{2cm} v=\frac{1}{y}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-fb8308850d90dc8397b289330e9e674c_l3.png)
La ecuación original quedaría
![Rendered by QuickLaTeX.com \left\{\begin{matrix} u^2+v^2=13\\ u-v=1 \ \ \ \end{matrix}\right.](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-9a312743909f1ebd080485815dd00bd8_l3.png)
2 Despejamos una incógnita de una de las ecuaciones
![Rendered by QuickLaTeX.com u=1+v](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-38f6c07d11d55a21da7c6c65dc35ac6f_l3.png)
3 Sustituimos en la otra ecuación
![Rendered by QuickLaTeX.com (1+v)^2+v^2=13](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-8c58e72a983b091849ff1e4f56b7d3dd_l3.png)
Desarrollamos
![Rendered by QuickLaTeX.com 1+2v+v^2+v^2=13](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-bd13c5afe4b10b224dc1ca15aac1838d_l3.png)
![Rendered by QuickLaTeX.com 2v^2+2v-12=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-5bd958f0bb256f0d7658f731b6e5df0a_l3.png)
![Rendered by QuickLaTeX.com v^2+v-6=0](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-a9765cc40fb92f3aad3d1fbfa5b61efc_l3.png)
4 Resolvemos
Por fórmula general sabemos que
![Rendered by QuickLaTeX.com \displaystyle v=\frac{-1\pm \sqrt{1+24}}{2}=\frac{-1\pm 5}{2}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-4b8b179228bf2c9658d582b06ee1ec29_l3.png)
![Rendered by QuickLaTeX.com v_1=2](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-bfba1b9da70a02f8084940d7b324d75a_l3.png)
![Rendered by QuickLaTeX.com v_2=-3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-2f926ba1bc142c6eb9e8f9610237220d_l3.png)
5 Obtenemos el valor de la otra incógnita
![Rendered by QuickLaTeX.com v=2 \hspace{2cm} u=1+2 \hspace{2cm} u=3](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-f3d4168510c18e6b57a981a2516a811c_l3.png)
![Rendered by QuickLaTeX.com v=-3 \hspace{2cm} u=1-3 \hspace{2cm} u=-2](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-18dc3b0cd1d5ff6adf67f6185fa4bc85_l3.png)
5 Consideramos el cambio de variable que hicimos al principio
Con la solución de
![Rendered by QuickLaTeX.com \displaystyle v=2=\frac{1}{y} \hspace{1cm} \Rightarrow \hspace{1cm} y=\frac{1}{2}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-eea621831a2f0906f11fb664c5a77d7b_l3.png)
![Rendered by QuickLaTeX.com \displaystyle u=3=\frac{1}{x} \hspace{1cm} \Rightarrow \hspace{1cm} x=\frac{1}{3}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-8b939d470dc968ffe8b72d1c8c47f756_l3.png)
Con la solución de
![Rendered by QuickLaTeX.com \displaystyle v=-3=\frac{1}{y} \hspace{1cm} \Rightarrow \hspace{1cm} y=-\frac{1}{3}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-530dd60dd0bba11e6fa4b9d2c5570218_l3.png)
![Rendered by QuickLaTeX.com \displaystyle u=-2=\frac{1}{x} \hspace{1cm} \Rightarrow \hspace{1cm} x=-\frac{1}{2}](https://www.superprof.es/apuntes/wp-content/ql-cache/quicklatex.com-e0f904f09208e2a32554707c563073d1_l3.png)