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15.5: Resolver ecuaciones racionales - Matemáticas


Después de definir los términos "expresión" y "ecuación" anteriormente, los hemos utilizado a lo largo de este libro. Ahora resolveremos un ecuación racional.

Debes asegurarte de conocer la diferencia entre expresiones racionales y ecuaciones racionales. La ecuación contiene un signo igual.

[ dfrac {1} {8} x + dfrac {1} {2} quad quad dfrac {1} {8} x + dfrac {1} {2} = dfrac {1} {4} sin número ]

[ dfrac {1} {n-3} + dfrac {1} {n + 4} quad quad quad quad dfrac {1} {n-3} + dfrac {1} {n + 4} = dfrac {15} {n ^ {2} + n-12} nonumber ]

Resolver ecuaciones racionales

Ya hemos resuelto ecuaciones lineales que contenían fracciones. Encontramos el MCD de todas las fracciones en la ecuación y luego multiplicamos ambos lados de la ecuación por el MCD para "borrar" las fracciones.

Usaremos la misma estrategia para resolver ecuaciones racionales. Multiplicaremos ambos lados de la ecuación por el LCD. Entonces, tendremos una ecuación que no contiene expresiones racionales y, por lo tanto, es mucho más fácil de resolver para nosotros. Pero debido a que la ecuación original puede tener una variable en un denominador, debemos tener cuidado de no terminar con una solución que haga que un denominador sea igual a cero.

Entonces, antes de comenzar a resolver una ecuación racional, la examinamos primero para encontrar los valores que harían cero cualquier denominador. De esa forma, cuando resolvemos una ecuación racional sabremos si hay alguna solución algebraica que debamos descartar.

Una solución algebraica de una ecuación racional que haría que cualquiera de las expresiones racionales no estuviera definida se llama solución extraña a una ecuación racional.

Solución extraña a una ecuación racional

Un solución extraña a una ecuación racional es una solución algebraica que haría que cualquiera de las expresiones de la ecuación original no estuviera definida.

Observamos cualquier posible solución extraña, (c ), escribiendo (x neq c ) junto a la ecuación.

Ejemplo ( PageIndex {1} ): Cómo resolver una ecuación racional

Resolver: [ dfrac {1} {x} + dfrac {1} {3} = dfrac {5} {6} nonumber ]

Solución

Paso 1. Anote cualquier valor de la variable que haría cero cualquier denominador.

Si (x = 0 ), entonces ( dfrac {1} {x} ) no está definido. Entonces escribiremos (x neq 0 ) al lado de la ecuación.

[ dfrac {1} {x} + dfrac {1} {3} = dfrac {5} {6}, x neq 0 nonumber ]

Paso 2. Encuentra el mínimo común denominador de todos los denominadores de la ecuación.

Encuentre el LCD de ( dfrac {1} {x} ), ( dfrac {1} {3} ) y ( dfrac {5} {6} )

La pantalla LCD es (6x ).

Paso 3. Borre las fracciones multiplicando ambos lados de la ecuación por el LCD.

Multiplica ambos lados de la ecuación por el MCD, (6x ).

[{ color {rojo} 6 x} cdot left ( dfrac {1} {x} + dfrac {1} {3} right) = { color {rojo} 6 x} cdot left ( dfrac {5} {6} right) nonumber ]

Usa la propiedad distributiva.

[{ color {rojo} 6 x} cdot dfrac {1} {x} + { color {rojo} 6 x} cdot dfrac {1} {3} = { color {rojo} 6 x } cdot left ( dfrac {5} {6} right) nonumber ]

Simplifica y fíjate, ¡no más fracciones!

[6 + 2 x = 5 x nonumber ]

Paso 4. Resuelve la ecuación resultante.

Simplificar.

[ begin {alineado} & 6 = 3 x & 2 = x end {alineado} nonumber ]

Paso 5. Cheque.

Si alguno de los valores encontrados en el Paso 1 son soluciones algebraicas, deséchelos. Verifique las soluciones restantes en la ecuación original.

No obtuvimos 0 como solución algebraica.

[ dfrac {1} {x} + dfrac {1} {3} = dfrac {5} {6} nonumber ]

Sustituimos (x = 2 ) en la ecuación original.

[ begin {alineado} frac {1} {2} + frac {1} {3} & overset {?} {=} frac {5} {6} frac {3} {6 } + frac {2} {6} & overset {?} {=} frac {5} {6} frac {5} {6} & = frac {5} {6} surd final {alineado} nonumber ]

La solución es (x = 2 )

Ejercicio ( PageIndex {1} )

Resolver: [ dfrac {1} {y} + dfrac {2} {3} = dfrac {1} {5} nonumber ]

Respuesta

(y = - dfrac {7} {15} )

Ejercicio ( PageIndex {2} )

Resolver: [ dfrac {2} {3} + dfrac {1} {5} = dfrac {1} {x} nonumber ]

Respuesta

(x = dfrac {13} {15} )

Se muestran los pasos de este método.

cómo resolver ecuaciones con expresiones racionales.

  • Paso 1. Anote cualquier valor de la variable que haría cero cualquier denominador.
  • Paso 2. Encuentra el mínimo común denominador de todos los denominadores en la ecuación.
  • Paso 3. Borre las fracciones multiplicando ambos lados de la ecuación por el LCD.
  • Paso 4. Resuelve la ecuación resultante.
  • Paso 5. Verificar:
    • Si alguno de los valores encontrados en el Paso 1 son soluciones algebraicas, deséchelos.
    • Verifique las soluciones restantes en la ecuación original.

Siempre comenzamos anotando los valores que harían que cualquier denominador sea cero.

Ejemplo ( PageIndex {2} ): Cómo resolver una ecuación racional usando la propiedad del producto cero

Resolver: [1- dfrac {5} {y} = - dfrac {6} {y ^ {2}} nonumber ]

Solución

Anote cualquier valor de la variable que haría cero cualquier denominador.

[1- dfrac {5} {y} = - dfrac {6} {y ^ {2}}, y neq 0 nonumber ]

Encuentra el mínimo común denominador de todos los denominadores en la ecuación. La pantalla LCD es (y ^ 2 ).

Borre las fracciones multiplicando ambos lados de la ecuación por el LCD.

[y ^ {2} left (1- dfrac {5} {y} right) = y ^ {2} left (- dfrac {6} {y ^ {2}} right) nonumber ]

Distribuir.

[y ^ {2} cdot 1-y ^ {2} left ( dfrac {5} {y} right) = y ^ {2} left (- dfrac {6} {y ^ {2 }} derecha) nonumber ]

Multiplicar.

[y ^ {2} -5 y = -6 nonumber ]

Resuelve la ecuación resultante. Primero escribe la ecuación cuadrática en forma estándar.

[y ^ {2} -5 y + 6 = 0 nonumber ]

Factor.

[(y-2) (y-3) = 0 nonumber ]

Utilice la propiedad de producto cero.

[y-2 = 0 text {o} y-3 = 0 nonumber ]

Resolver.

[y = 2 text {o} y = 3 nonumber ]

Cheque. No obtuvimos (0 ) como solución algebraica.

Marca (y = 2 ) y (y = 3 ) en la ecuación original.

[1- dfrac {5} {y} = - dfrac {6} {y ^ {2}} quad quad quad 1- dfrac {5} {y} = - dfrac {6} { y ^ {2}} nonumber ]

[1- dfrac {5} {2} overset {?} {=} - dfrac {6} {2 ^ {2}} quad quad quad 1- dfrac {5} {3} desbordado {?} {=} - dfrac {6} {3 ^ {2}} nonumber ]

[1- dfrac {5} {2} overset {?} {=} - dfrac {6} {4} quad quad quad 1- dfrac {5} {3} overset {?} {=} - dfrac {6} {9} nonumber ]

[ dfrac {2} {2} - dfrac {5} {2} overset {?} {=} - dfrac {6} {4} quad quad quad dfrac {3} {3} - dfrac {5} {3} overset {?} {=} - dfrac {6} {9} nonumber ]

[- dfrac {3} {2} overset {?} {=} - dfrac {6} {4} quad quad quad - dfrac {2} {3} overset {?} {= } - dfrac {6} {9} nonumber ]

[- dfrac {3} {2} = - dfrac {3} {2} surd quad quad quad - dfrac {2} {3} = - dfrac {2} {3} surd sin número ]

La solución es (y = 2, y = 3 )

Ejercicio ( PageIndex {3} )

Resolver: [1- dfrac {2} {x} = dfrac {15} {x ^ {2}} nonumber ]

Respuesta

(x = -3, x = 5 )

Ejercicio ( PageIndex {4} )

Resolver: [1- dfrac {4} {y} = dfrac {12} {y ^ {2}} nonumber ]

Respuesta

(y = -2, y = 6 )

En el siguiente ejemplo, el último denominador es una diferencia de cuadrados. Recuerde factorizarlo primero para encontrar la pantalla LCD.

Ejemplo ( PageIndex {3} )

Resolver: [ dfrac {2} {x + 2} + dfrac {4} {x-2} = dfrac {x-1} {x ^ {2} -4} nonumber ]

Solución

Tenga en cuenta cualquier valor de la variable que haría cero cualquier denominador.

[ dfrac {2} {x + 2} + dfrac {4} {x-2} = dfrac {x-1} {(x + 2) (x-2)}, x neq-2, x neq 2 nonumber ]

Encuentra el mínimo común denominador de todos los denominadores en la ecuación. El LCD es ((x + 2) (x-2) ).

Borre las fracciones multiplicando ambos lados de la ecuación por el LCD.

[(x + 2) (x-2) left ( dfrac {2} {x + 2} + dfrac {4} {x-2} right) = (x + 2) (x-2) left ( dfrac {x-1} {x ^ {2} -4} right) nonumber ]

Distribuir.

[(x + 2) (x-2) dfrac {2} {x + 2} + (x + 2) (x-2) dfrac {4} {x-2} = (x + 2) ( x-2) left ( dfrac {x-1} {x ^ {2} -4} right) nonumber ]

Elimina los factores comunes.

[ cancelar {(x + 2)} (x-2) dfrac {2} { cancelar {x + 2}} + (x + 2) { cancelar {(x-2)}} dfrac { 4} { cancel {x-2}} = cancel {(x + 2) (x-2)} left ( dfrac {x-1} { cancel {x ^ {2} -4}} derecha) nonumber ]

Simplificar.

[2 (x-2) +4 (x + 2) = x-1 nonumber ]

Distribuir.

[2 x-4 + 4 x + 8 = x-1 nonumber ]

Resolver.

[ begin {alineado} 6 x + 4 & = x-1 5 x & = - 5 x & = - 1 end {alineado} ]

Compruebe: No obtuvimos 2 o −2 como soluciones algebraicas.

Marca (x = -1 ) en la ecuación original.

[ begin {alineado} dfrac {2} {x + 2} + dfrac {4} {x-2} & = dfrac {x-1} {x ^ {2} -4} dfrac {2} {(- 1) +2} + dfrac {4} {(- 1) -2} & overset {?} {=} Dfrac {(- 1) -1} {(- 1) ^ {2} -4} dfrac {2} {1} + dfrac {4} {- 3} & overset {?} {=} Dfrac {-2} {- 3} dfrac { 6} {3} - dfrac {4} {3} & overset {?} {=} Dfrac {2} {3} dfrac {2} {3} & = dfrac {2} {3 } surd end {alineado} nonumber ]

La solución es (x = -1 ).

Ejercicio ( PageIndex {5} )

Resolver: [ dfrac {2} {x + 1} + dfrac {1} {x-1} = dfrac {1} {x ^ {2} -1} nonumber ]

Respuesta

(x = dfrac {2} {3} )

Ejercicio ( PageIndex {6} )

Resolver: [ dfrac {5} {y + 3} + dfrac {2} {y-3} = dfrac {5} {y ^ {2} -9} nonumber ]

Respuesta

(y = 2 )

En el siguiente ejemplo, el primer denominador es un trinomio. Recuerde factorizarlo primero para encontrar la pantalla LCD.

Ejemplo ( PageIndex {4} )

Resolver: [ dfrac {m + 11} {m ^ {2} -5 m + 4} = dfrac {5} {m-4} - dfrac {3} {m-1} nonumber ]

Solución

Anote cualquier valor de la variable que haría cero cualquier denominador. Usa la forma factorizada del denominador cuadrático.

[ dfrac {m + 11} {(m-4) (m-1)} = dfrac {5} {m-4} - dfrac {3} {m-1}, m neq 4, m neq 1 nonumber ]

Encuentra el mínimo común denominador de todos los denominadores en la ecuación. La pantalla LCD es ((m-4) (m-1) )

Borre las fracciones multiplicando ambos lados de la ecuación por el LCD.

[(m-4) (m-1) left ( dfrac {m + 11} {(m-4) (m-1)} right) = (m-4) (m-1) left ( dfrac {5} {m-4} - dfrac {3} {m-1} right) nonumber ]

Distribuir.

[(m-4) (m-1) left ( dfrac {m + 11} {(m-4) (m-1)} right) = (m-4) (m-1) dfrac {5} {m-4} - (m-4) (m-1) dfrac {3} {m-1} nonumber ]

Elimina los factores comunes.

[ cancel {(m-4) (m-1)} left ( dfrac {m + 11} { cancel {(m-4) (m-1)}} right) = cancel {( m-4)} (m-1) dfrac {5} { cancel {m-4}} - (m-4) cancel {(m-1)} dfrac {3} { cancel {m- 1}} nonumber ]

Simplificar.

[m + 11 = 5 (m-1) -3 (m-4) nonumber ]

Resuelve la ecuación resultante.

[ begin {alineado} m + 11 & = 5 m-5-3 m + 12 4 & = m end {alineado} nonumber ]

Cheque. La única solución algebraica era 4, pero dijimos que 4 haría un denominador igual a cero. La solución algebraica es una solución extraña.

No hay solución para esta ecuación.

Ejercicio ( PageIndex {7} )

Resolver: [ dfrac {x + 13} {x ^ {2} -7 x + 10} = dfrac {6} {x-5} - dfrac {4} {x-2} nonumber ]

Respuesta

No hay solución.

Ejercicio ( PageIndex {8} )

Resolver: [ dfrac {y-6} {y ^ {2} +3 y-4} = dfrac {2} {y + 4} + dfrac {7} {y-1} nonumber ]

Respuesta

No hay solución.

La ecuación que resolvimos en el ejemplo anterior tenía solo una solución algebraica, pero era una solución extraña. Eso nos dejó sin solución a la ecuación. En el siguiente ejemplo obtenemos dos soluciones algebraicas. Aquí una o ambas podrían ser soluciones extrañas.

Ejemplo ( PageIndex {5} )

Resolver: [ dfrac {y} {y + 6} = dfrac {72} {y ^ {2} -36} +4 nonumber ]

Solución

Factoriza todos los denominadores, de modo que podamos anotar cualquier valor de la variable que haría que cualquier denominador sea cero.

[ dfrac {y} {y + 6} = dfrac {72} {(y-6) (y + 6)} + 4, y neq 6, y neq-6 nonumber ]

Encuentra el mínimo denominador común. El LCD es ((y-6) (y + 6) )

Limpia las fracciones.

[(y-6) (y + 6) left ( dfrac {y} {y + 6} right) = (y-6) (y + 6) left ( dfrac {72} {(y -6) (y + 6)} + 4 right) nonumber ]

Simplificar.

[(y-6) cdot y = 72 + (y-6) (y + 6) cdot 4 nonumber ]

Simplificar.

[y (y-6) = 72 + 4 left (y ^ {2} -36 right) nonumber ]

Resuelve la ecuación resultante.

[ begin {align} y ^ {2} -6 y & = 72 + 4 y ^ {2} -144 0 & = 3 y ^ {2} +6 y-72 0 & = 3 left (y ^ {2} +2 y-24 right) 0 & = 3 (y + 6) (y-4) y & = - 6, y = 4 end {alineado} nonumber ]

Cheque.

(y = -6 ) es una solución extraña. Marca (y = 4 ) en la ecuación original.

[ begin {alineado} dfrac {y} {y + 6} & = dfrac {72} {y ^ {2} -36} +4 dfrac {4} {4 + 6} & overset {?} {=} dfrac {72} {4 ^ {2} -36} +4 dfrac {4} {10} & overset {?} {=} dfrac {72} {- 20} +4 dfrac {4} {10} & overset {?} {=} - dfrac {36} {10} + dfrac {40} {10} dfrac {4} {10} & = dfrac {4} {10} surd end {alineado} nonumber ]

La solución es (y = 4 ).

Ejercicio ( PageIndex {9} )

Resolver: [ dfrac {x} {x + 4} = dfrac {32} {x ^ {2} -16} +5 nonumber ]

Respuesta

(x = 3 )

Ejercicio ( PageIndex {10} )

Resolver: [ dfrac {y} {y + 8} = dfrac {128} {y ^ {2} -64} +9 nonumber ]

Respuesta

(y = 7 )

En algunos casos, todas las soluciones algebraicas son extrañas.

Ejemplo ( PageIndex {6} )

Resolver: [ dfrac {x} {2 x-2} - dfrac {2} {3 x + 3} = dfrac {5 x ^ {2} -2 x + 9} {12 x ^ {2} -12} nonumber ]

Solución

Comenzaremos factorizando todos los denominadores, para facilitar la identificación de soluciones extrañas y el LCD.

[ dfrac {x} {2 (x-1)} - dfrac {2} {3 (x + 1)} = dfrac {5 x ^ {2} -2 x + 9} {12 (x- 1) (x + 1)} nonumber ]

Anote cualquier valor de la variable que haría cero cualquier denominador.

[ dfrac {x} {2 (x-1)} - dfrac {2} {3 (x + 1)} = dfrac {5 x ^ {2} -2 x + 9} {12 (x- 1) (x + 1)}, x neq 1, x neq-1 nonumber ]

Encuentra el mínimo denominador común. El LCD es (12 (x-1) (x + 1) ).

Limpia las fracciones.

[12 (x-1) (x + 1) left ( dfrac {x} {2 (x-1)} - dfrac {2} {3 (x + 1)} right) = 12 (x -1) (x + 1) left ( dfrac {5 x ^ {2} -2 x + 9} {12 (x-1) (x + 1)} right) nonumber ]

Simplificar.

[6 (x + 1) cdot x-4 (x-1) cdot 2 = 5 x ^ {2} -2 x + 9 nonumber ]

Simplificar.

[6 x (x + 1) -4 cdot 2 (x-1) = 5 x ^ {2} -2 x + 9 nonumber ]

Resuelve la ecuación resultante.

[ begin {align} 6 x ^ {2} +6 x-8 x + 8 & = 5 x ^ {2} -2 x + 9 x ^ {2} -1 & = 0 (x-1 ) (x + 1) & = 0 x & = 1 text {o} x = -1 end {alineado} nonumber ]

Cheque.

(x = 1 ) y (x = -1 ) son soluciones extrañas.

La ecuación no tiene solución.

Ejercicio ( PageIndex {11} )

Resolver: [ dfrac {y} {5 y-10} - dfrac {5} {3 y + 6} = dfrac {2 y ^ {2} -19 y + 54} {15 y ^ {2} -60} nonumber ]

Respuesta

No hay solución.

Ejercicio ( PageIndex {12} )

Resolver: [ dfrac {z} {2 z + 8} - dfrac {3} {4 z-8} = dfrac {3 z ^ {2} -16 z-16} {8 z ^ {2} +2 z-64} nonumber ]

Respuesta

No hay solución.

Ejemplo ( PageIndex {7} )

Resolver: [ dfrac {4} {3 x ^ {2} -10 x + 3} + dfrac {3} {3 x ^ {2} +2 x-1} = dfrac {2} {x ^ {2} -2 x-3} nonumber ]

Solución

Factoriza todos los denominadores, de modo que podamos anotar cualquier valor de la variable que haría que cualquier denominador sea cero.

[ dfrac {4} {(3 x-1) (x-3)} + dfrac {3} {(3 x-1) (x + 1)} = dfrac {2} {(x-3 ) (x + 1)}, x neq-1, x neq dfrac {1} {3}, x neq 3 nonumber ]

Encuentra el mínimo denominador común. El LCD es ((3 x-1) (x + 1) (x-3) ).

Limpia las fracciones.

[(3 x-1) (x + 1) (x-3) left ( dfrac {4} {(3 x-1) (x-3)} + dfrac {3} {(3 x- 1) (x + 1)} right) = (3 x-1) (x + 1) (x-3) left ( dfrac {2} {(x-3) (x + 1)} right ) sin número ]

Simplificar.

[4 (x + 1) +3 (x-3) = 2 (3 x-1) nonumber ]

Distribuir.

[4 x + 4 + 3 x-9 = 6 x-2 nonumber ]

Simplificar.

[7 x-5 = 6 x-2 nonumber ]

[x = 3 nonumber ]

La única solución algebraica fue (x = 3 ), pero dijimos que (x = 3 ) haría un denominador igual a cero. La solución algebraica es una solución extraña.

No hay solución para esta ecuación.

Ejercicio ( PageIndex {13} )

Resolver: [ dfrac {15} {x ^ {2} + x-6} - dfrac {3} {x-2} = dfrac {2} {x + 3} nonumber ]

Respuesta

No hay solución.

Ejercicio ( PageIndex {14} )

Resolver: [ dfrac {5} {x ^ {2} +2 x-3} - dfrac {3} {x ^ {2} + x-2} = dfrac {1} {x ^ {2} +5 x + 6} nonumber ]

Respuesta

No hay solución.

Utilice funciones racionales

Trabajar con funciones que están definidas por expresiones racionales a menudo conduce a ecuaciones racionales. Nuevamente, usamos las mismas técnicas para resolverlos.

Ejemplo ( PageIndex {8} )

Para una función racional, (f (x) = dfrac {2 x-6} {x ^ {2} -8 x + 15} ):

  1. Encuentra el dominio de la función
  2. Resuelve (f (x) = 1 )
  3. Encuentra los puntos en la gráfica en este valor de función.

Solución

  1. El dominio de una función racional son todos los números reales excepto aquellos que hacen que la expresión racional sea indefinida. Entonces, para encontrarlos, igualaremos el denominador a cero y resolveremos.

[ begin {align} x ^ {2} -8 x + 15 & = 0 (x-3) (x-5) & = 0 quad text {Factoriza el trinomio.} x-3 & = 0 quad text {Utilice la propiedad del producto cero.} x-5 & = 0 quad text {Utilice la propiedad del producto cero.} x = 3 & ; x = 5 text {Resolver.} end {alineado} nonumber ]

El dominio son todos los números reales excepto (x neq 3, x neq 5 )

  1. [f (x) = 1 nonumber ]

Sustituir en la expresión racional.

[ dfrac {2 x-6} {x ^ {2} -8 x + 15} = 1 nonumber ]

Factoriza el denominador.

[ dfrac {2 x-6} {(x-3) (x-5)} = 1 nonumber ]

Multiplica ambos lados por el MCD, ((x-3) (x-5) )

[(x-3) (x-5) left ( dfrac {2 x-6} {(x-3) (x-5)} right) = (x-3) (x-5) ( 1) nonumber ]

Simplificar.

[2 x-6 = x ^ {2} -8 x + 15 nonumber ]

Resolver.

[0 = x ^ {2} -10 x + 21 nonumber ]

Factor.

[0 = (x-7) (x-3) nonumber ]

Utilice la propiedad de producto cero.

[x-7 = 0 quad x-3 = 0 nonumber ]

Resolver.

[x = 7 quad x = 3 nonumber ]

  1. El valor de la función es 1 cuando (x = 7, x = 3 ). Entonces, los puntos en la gráfica de esta función cuando (f (x) = 1 ), será ((7,1), (3,1) ).

Ejercicio ( PageIndex {15} )

Para una función racional, (f (x) = dfrac {8-x} {x ^ {2} -7 x + 12} )

  1. Encuentra el dominio de la función.
  2. Resuelve (f (x) = 3 ).
  3. Encuentra los puntos en la gráfica en este valor de función.
Respuesta
  1. El dominio son todos los números reales excepto (x neq 3 ) y (x neq 4 )
  2. (x = 2, x = dfrac {14} {3} )
  3. ((2,3), left ( dfrac {14} {3}, 3 right) )

Ejercicio ( PageIndex {16} )

Para una función racional, (f (x) = dfrac {x-1} {x ^ {2} -6 x + 5} )

  1. Resuelve (f (x) = 4 ).
  2. Encuentra los puntos en la gráfica en este valor de función.
Respuesta
  1. El dominio son todos los números reales excepto (x neq 1 ) y (x neq 5 )
  2. (x = dfrac {21} {4} )
  3. ( izquierda ( dfrac {21} {4}, 4 derecha) )

Resolver una ecuación racional para una variable específica

Cuando resolvimos ecuaciones lineales, aprendimos cómo resolver una fórmula para una variable específica. Muchas fórmulas utilizadas en negocios, ciencia, economía y otros campos usan ecuaciones racionales para modelar la relación entre dos o más variables. Ahora veremos cómo resolver una ecuación racional para una variable específica.

Cuando desarrollamos la fórmula punto-pendiente a partir de nuestra fórmula de pendiente, aclaramos las fracciones multiplicando por el MCD.

[ begin {alineado} m & = frac {y-y_ {1}} {x-x_ {1}} m left (x-x_ {1} right) & = left ( frac {y-y_ {1}} {x-x_ {1}} right) left (x-x_ {1} right) quad text {Multiplica ambos lados de la ecuación por} x-x_1. m left (x-x_ {1} right) & = y-y_ {1} quad text {Simplificar.} y-y_ {1} & = m left (x-x_ {1} right) quad text {Reescribe la ecuación con los términos y a la izquierda.} end {alineado} nonumber ]

En el siguiente ejemplo, usaremos la misma técnica con la fórmula de pendiente que usamos para obtener la forma punto-pendiente de una ecuación de una línea que pasa por el punto ((2,3) ). Agregaremos un paso más para resolver (y ).

Ejemplo ( PageIndex {9} )

Resuelve: (m = dfrac {y-2} {x-3} ) para (y ).

Solución

[m = dfrac {y-2} {x-3} nonumber ]

Anote cualquier valor de la variable que haría cero cualquier denominador.

[m = dfrac {y-2} {x-3}, x neq 3 nonumber ]

Limpia las fracciones multiplicando ambos lados de la ecuación por el MCD, (x-3 ).

[(x-3) m = (x-3) left ( dfrac {y-2} {x-3} right) nonumber ]

Simplificar.

[x m-3 m = y-2 nonumber ]

Aislar el término con (y ).

[x m-3 m + 2 = y nonumber ]

Ejercicio ( PageIndex {17} )

Resuelve: (m = dfrac {y-5} {x-4} ) para (y ).

Respuesta

(y = metro x-4 metro + 5 )

Ejercicio ( PageIndex {18} )

Resuelve: (m = dfrac {y-1} {x + 5} ) para (y ).

Respuesta

(y = metro x + 5 metros + 1 )

Recuerde multiplicar ambos lados por el LCD en el siguiente ejemplo.

Ejemplo ( PageIndex {10} )

Resuelve: ( dfrac {1} {c} + dfrac {1} {m} = 1 ) para (c )

Solución

[ dfrac {1} {c} + dfrac {1} {m} = 1 text {para} c nonumber ]

Anote cualquier valor de la variable que haría cero cualquier denominador.

[ dfrac {1} {c} + dfrac {1} {m} = 1, c neq 0, m neq 0 nonumber ]

Borre las fracciones multiplicando ambos lados de las ecuaciones por el MCD, (cm ).

[cm left ( dfrac {1} {c} + dfrac {1} {m} right) = cm (1) nonumber ]

Distribuir.

[cm left ( frac {1} {c} right) + cm frac {1} {m} = cm (1) nonumber ]

Simplificar.

[m + c = cm nonumber ]

Recopile los términos con (c ) a la derecha.

[m = cm-c nonumber ]

Factoriza la expresión de la derecha.

[m = c (m-1) nonumber ]

Para aislar (c ), divide ambos lados entre (m-1 ).

[ dfrac {m} {m-1} = dfrac {c (m-1)} {m-1} nonumber ]

Simplifique eliminando factores comunes.

[ dfrac {m} {m-1} = c nonumber ]

Observe que aunque excluimos (c = 0 ) y (m = 0 ) de la ecuación original, ahora también debemos afirmar que (m neq 1 ).

Ejercicio ( PageIndex {19} )

Resuelve: ( dfrac {1} {a} + dfrac {1} {b} = c ) para (a ).

Respuesta

(a = dfrac {b} {c b-1} )

Ejercicio ( PageIndex {20} )

Resolver: ( dfrac {2} {x} + dfrac {1} {3} = dfrac {1} {y} ) para (y )

Respuesta

(y = dfrac {3 x} {x + 6} )


Resolver ecuaciones racionales

En el mundo de los negocios y los campos de la tecnología, hay momentos en que las matemáticas modelan una situación del mundo real con una ecuación racional. Para resolver estas ecuaciones, necesitamos entender cómo manipular estas ecuaciones, agregando partes o multiplicando expresiones racionales.

A veces, la ecuación se parecerá a una proporción. Cuando este sea el caso, resuelva la ecuación mediante multiplicación cruzada.

Ejemplo 1: multiplicación cruzada

5 x + 1 = 10 x - 1
5 (x - 1) = 10 (x + 1) Multiplica en cruz.

5x - 5 = 10x + 10 Usa la propiedad distributiva.

-5x = 15 Resta 10x y suma 5 en cada lado.

x = -3 Divide ambos lados entre -5.

Cuando un lado de la ecuación tiene suma o resta con razones, resuelva multiplicando cada término por el mínimo denominador común (MCD).

Ejemplo 2: multiplicar por la pantalla LCD

4 x & # 8226 (8 x + 14 4) = (4) & # 8226 4 x Multiplica ambos lados por la pantalla LCD.

32 + 14x = 16x Usa la propiedad distributiva.

32 = 2x Resta ambos lados por 14x.

16 = x Divide ambos lados entre 2.

A menudo, al multiplicar la pantalla LCD, habrá dos soluciones.

Ejemplo 3: más de una solución

9 x + 3 + x - 3 x - 1 = 4 x 2 + 2 x - 3

(x + 3) (x - 1) (9 x + 3 + x - 3 x - 1) = (4 x 2 + 2 x - 3) (x + 3) (x - 1) Multiplica por el LCD.

(x - 1) 9 + (x - 3) (x + 3) = 4 Elimina términos comunes en los denominadores.

9x - 9 + x 2-9 = 4 Multiplica los binomios.

x 2 + 9x - 22 = 0 Combinar términos semejantes y restar 4 en ambos lados.

Hay momentos en que las soluciones son extrañas. Cuando esto sucede, una de las soluciones devolverá un denominador que es igual a cero.

Ejemplo 4: Soluciones extrañas

6 x - 5 = 8 x 2 x 2 - 25 - 4 x x + 5

(x - 5) (x + 5) (6 x - 5) = (8 x 2 x 2 - 25 - 4 x x + 5) (x - 5) (x + 5) Multiplica por el MCD.

(x + 5) 6 = 8x 2 - 4x (x - 5) Elimina términos comunes en los denominadores.

6x + 30 = 8x 2 - 4x 2 + 20x Usa la propiedad distributiva.

4x 2 + 14x - 30 = 0 Mover todos los términos al lado izquierdo.

x = & # x00A0 3 2, o x = -5 x = -5 no puede ser una solución, ya que la ecuación no está definida en esta coordenada.

Ejemplo 5: promedio de bateo

En los primeros 4 meses de la temporada de béisbol, Tommy bateó 20 de 85 turnos al bate. ¿Cuántos hits consecutivos necesita hacer para tener un promedio de bateo de 0.350?

Solución: El promedio de bateo es el número de hits dividido por el número de turnos al bate.

El promedio actual de Tommy es 20/85 = 0.235

Para que el promedio de Tommy aumente a 0.350, resuelva la ecuación 0.35 = 20 + x 85 + x

Resuelve por multiplicación cruzada.

0.35 = 20 + x 85 + x Configure el promedio.

0.35 (85 + x) = 20 + x Multiplica en cruz.

29,75 + 0,35x = 20 + x Usa la propiedad de distribución.

9,75 = 0,65x restar 20 y 0,35x en ambos lados

Tommy necesita batear 15 bates consecutivos para elevar su promedio de bateo a 0.350

Ejemplo 6: Trabajar juntos

Sue y Ann trabajan juntas para limpiar casas. Sue tarda 7 horas en limpiar una casa. Con la ayuda de Ann, solo les toma 3 horas limpiar. ¿Cuánto tiempo le toma a Ann limpiar la misma casa sola?

La mejor manera de responder a este escenario es sumar la cantidad de trabajo que se realiza por hora.

Sue limpia 1/7 de la casa por hora.

Ann limpia 1 / A de la casa por hora.

Sue y Ann limpian 1/3 de una casa en una hora.

21 A (1 7 + 1 A) = (1 3) 21 A Multiplique por el LCD.

3A + 21 = 7A Usa la propiedad distributiva.

21 = 4A Restar 3A en ambos lados

Ann tarda 5 horas y 15 minutos en limpiar la casa.

Para vincular a esto Resolver ecuaciones racionales página, copie el siguiente código en su sitio:


Álgebra: resolución de ecuaciones racionales

¿Alguna vez has notado que muchos matemáticos no tienen grandes habilidades sociales? (Si no te has dado cuenta, debe ser porque no conoces a muchos matemáticos.) Hay un viejo chiste que dice: "¿Cómo puedes identificar a un matemático extrovertido? tu zapatos en lugar de los suyos ".

¿Por qué algunas personas de matemáticas no pueden lidiar con el mundo exterior? ¿Es la brillante luz del día que quema a sus alumnos detrás de sus gruesos anteojos, o los calzones que soportaron en la escuela? No creo que ninguno de los dos tenga la culpa. Creo que es porque están acostumbrados a un mundo total y totalmente bajo su control. Verá, en el mundo de las matemáticas, si no le gusta algo, puede manipular las reglas del universo y hacer desaparecer cualquier disgusto.

Precauciones de Kelley

Si multiplica una ecuación por algo que contiene un X, es posible que esté agregando soluciones incorrectas. Siempre verifique para asegurarse de que cualquier respuesta que obtenga se pueda conectar a la ecuación original. (En otras palabras, asegúrese de que ninguno de los denominadores se convierta en 0 cuando ingrese sus respuestas para X.)

Caso en cuestión: a la mayoría de los matemáticos no les gustan las fracciones. No es porque no los entiendan, es solo que la necesidad constante de denominadores comunes es molesta. Por lo tanto, siempre que surja la oportunidad, la gente de matemáticas eliminará por completo las fracciones de su paisaje. Por ejemplo, puede eliminar muy fácilmente cada fracción de una ecuación simplemente multiplicando todo en esa ecuación por su mínimo común denominador.

Ejemplo 1: Resuelve la ecuación.

Solución: Esta ecuación contiene tres expresiones racionales, su objetivo será eliminar todas esas fracciones para hacer la ecuación mucho más simple de resolver. Empiece por factorizar cualquier expresión que pueda en la ecuación. (En este caso, la cuadrática se puede factorizar).

El mínimo común denominador de las tres fracciones es (X + 5)(X + 2). Si multiplicas el ecuación completa por esa expresión, las fracciones desaparecerán. (Técnicamente hablando, multiplicarás la ecuación por (X + 5)(X + 2) 1, que es exactamente la misma expresión, solo te recuerda que debes multiplicar el mínimo común denominador por el numerador de cada fracción individual sin cambiar los denominadores).

Puedes simplificar todas esas fracciones.

Observe que cada uno de esos denominadores ha sido eliminado, y técnicamente ahora son todos iguales a 1. Sin embargo, no es necesario escribir un denominador de 1, por lo que puede reescribir la ecuación usando solo los numeradores.

Distribuya el X en el segundo término y combine todos los términos semejantes (estableciendo la ecuación igual a 0).

  • X + 2 + X 2 + 5X = 2X - 1
  • X 2 + 6X + 2 = 2X - 1
  • X 2 + 4X + 3 = 0

¡Oye! Queda una vieja ecuación cuadrática que puedes resolver factorizando.

Si conecta ambas soluciones en la ecuación original, obtendrá afirmaciones verdaderas, por lo que ambas son respuestas válidas.

Tienes problemas

Problema 1: resuelve la ecuación X + 3 X - 8 + X X 2 - 6X - 16 = 1.

Extraído de The Complete Idiot's Guide to Algebra 2004 por W. Michael Kelley. Todos los derechos reservados, incluido el derecho de reproducción total o parcial en cualquier forma. Usado por acuerdo con Libros Alfa, miembro de Penguin Group (USA) Inc.


Resolver ecuaciones racionales

Estas lecciones ayudan a los estudiantes de Álgebra y 9º grado a aprender a resolver ecuaciones racionales.

La siguiente figura muestra cómo resolver ecuaciones racionales. Desplácese hacia abajo en la página para ver más ejemplos y soluciones.


Resolver ecuaciones racionales, conceptos básicos

Paso 1: multiplicar por LCD
Paso 2: resuelve la ecuación resultante
Paso 3: Verifique la solución. Rechaza cualquier solución que haga que el denominador = 0

Ejemplos:
Resuelve y comprueba la respuesta:
19 años / 6 - 5 años / 2 = 4
b / 5 - (b - 1) / 10 = 9/10

Ecuaciones racionales - Ayuda de álgebra

Los estudiantes aprenden que al resolver ecuaciones racionales, el primer paso es factorizar cada uno de los denominadores, si es posible, luego multiplicar ambos lados de la ecuación por el denominador común de todas las fracciones para deshacerse de las fracciones, y resolver desde aquí. .

Finalmente, verifique cada solución para ver si hace que un denominador en la ecuación original sea igual a cero. Si es así, entonces no puede ser una solución a la ecuación.

Ejemplo:
5 / (y + 2) = 13 (y 2 - 3y - 10) + (-3) / (y - 5)

Resolver ecuaciones racionales

Ejemplos:
Resuelve las siguientes ecuaciones, busca soluciones extrañas.

  1. 4 / x + 5/2 = -11 / x
  2. 5x / (x - 2) = 7 + 10 / (x - 2)
  3. (3x - 2) / (x - 2) = 6 / (x 2 - 4) + 1
  4. 2 / (x 2 - x) = 1 / (x - 1)
  5. 3 / (x + 2) = 6 / (x - 1)

Pruebe la calculadora Mathway gratuita y el solucionador de problemas a continuación para practicar varios temas matemáticos. Pruebe los ejemplos dados o escriba su propio problema y verifique su respuesta con las explicaciones paso a paso.

Agradecemos sus comentarios, comentarios y preguntas sobre este sitio o página. Envíe sus comentarios o consultas a través de nuestra página de Comentarios.


Recursos abiertos para álgebra de colegios comunitarios

Abrimos esta sección mirando hacia atrás en el ejemplo 12.3.2. Julia lleva a su familia a dar un paseo en barco a (12 ) millas río abajo y de regreso. El río fluye a una velocidad de (2 ) millas por hora y ella quiere conducir el bote a una velocidad constante, (v ) millas por hora corriente abajo y de regreso corriente arriba. Debido a la corriente del río, la velocidad real de viaje es de (v + 2 ) millas por hora corriente abajo y (v-2 ) millas por hora corriente arriba. Si Julia planea pasar (8 ) horas durante todo el viaje, ¿qué tan rápido debe conducir el bote?

El tiempo que le toma a Julia conducir el barco río abajo es ( frac <12>) horas, y aguas arriba es ( frac <12>) horas. La función para modelar el tiempo de todo el viaje es

donde (t ) representa el tiempo en horas. El viaje tomará (8 ) horas, por lo que queremos que (t (v) ) sea igual a (8 text <,> ) y tenemos:

En lugar de usar la gráfica de la función, resolveremos esta ecuación algebraicamente. Es posible que desee revisar la técnica de eliminación de denominadores discutida en la Subsección 2.3.2. Podemos usar la misma técnica con expresiones variables en los denominadores. Para eliminar las fracciones en esta ecuación, multiplicaremos ambos lados de la ecuación por el mínimo denominador común ((v-2) (v + 2) text <,> ) y tenemos:

Observación 12.5.2.

En este punto, lógicamente todo lo que sabemos es que el único posible las soluciones son (- 1 ) y (4 text <.> ) Debido al paso en el que se cancelaron los factores, es posible que en realidad no sean soluciones de la ecuación original. Cada uno de ellos podría ser lo que se llama un. Una solución extraña es un número que parecería ser una solución basada en el proceso de resolución, pero en realidad no hace que la ecuación original sea verdadera. Por ello, es importante que se comprueben estas soluciones propuestas. Tenga en cuenta que no estamos verificando si cometimos un error de cálculo, sino que estamos verificando si las soluciones propuestas realmente resuelven la ecuación original.

Algebraicamente, ambos valores parecen ser soluciones. En el contexto de este escenario, la velocidad del barco no puede ser negativa, por lo que solo tomamos la solución (4 text <.> ) Si Julia conduce a (4 ) millas por hora, todo el viaje tomaría (8 horas. Este resultado coincide con la solución del ejemplo 12.3.2.

Definición 12.5.3. Ecuación racional.

Una ecuación racional es una ecuación que involucra una o más expresiones racionales. Por lo general, consideramos que son ecuaciones que tienen la variable en el denominador de al menos un término.

Veamos otro problema de aplicación.

Ejemplo 12.5.4.

Se necesitan Ku (3 ) horas para pintar una habitación y Jacob (6 ) horas para pintar la misma habitación. Si trabajan juntos, ¿cuánto tiempo les tomaría pintar la habitación?

Como a Ku (3 ) horas le toma pintar la habitación, pinta ( frac <1> <3> ) de la habitación cada hora. Del mismo modo, Jacob pinta ( frac <1> <6> ) de la habitación cada hora. Si trabajan juntos, pintan ( frac <1> <3> + frac <1> <6> ) de la habitación cada hora.

Suponga que se necesitan (x ) horas para pintar la habitación si Ku y Jacob trabajan juntos. Esto implica que pintan ( frac <1>) de la habitación juntos cada hora. Ahora podemos escribir esta ecuación:

Para eliminar los denominadores, multiplicamos ambos lados de la ecuación por el denominador común de (3 text <,> ) (6 ) y (x text <,> ) que es (6x text <:> )

¿La posible solución (x = 2 ) se verifica como una solución real?

Lo hace, por lo que es una solución. Si Ku y Jacob trabajan juntos, les tomaría (2 ) horas pintar la habitación.

Estamos listos para delinear un proceso general para resolver una ecuación racional.

Proceso 12.5.5. Resolver ecuaciones racionales.

Para resolver una ecuación racional,

Encuentra el mínimo común denominador para todos los términos de la ecuación.

Multiplicar cada termino en la ecuación por el mínimo común denominador

Cada denominador debe cancelarse dejando un tipo de ecuación más simple para resolver. Usa el método anterior para resolver esa ecuación.

Veamos algunos ejemplos más de resolución de ecuaciones racionales.

Ejemplo 12.5.6.

El denominador común es (y (y + 1) text <.> ) Multiplicaremos ambos lados de la ecuación por (y (y + 1) text <:> )

¿La posible solución (y = -3 ) se verifica como una solución real?

Verifica, entonces (- 3 ) es una solución. Escribimos el conjunto de soluciones como ( <- 3 > text <.> )

Ejemplo 12.5.7.

El denominador común es (z-4 text <.> ) Multiplicaremos ambos lados de la ecuación por (z-4 text <:> )

¿Las posibles soluciones (z = 1 ) y (z = 4 ) se verifican como soluciones reales?

La posible solución (z = 4 ) en realidad no funciona, ya que conduce a la división por (0 ) en la ecuación. Es una solución extraña. Sin embargo, (z = 1 ) es una solución válida. La única solución a la ecuación es (1 text <,> ) y así podemos escribir el conjunto de soluciones como ( <1 > text <.> )

Ejemplo 12.5.8.

Para encontrar el denominador común, necesitamos factorizar todos los denominadores si es posible:

Ahora podemos ver que el denominador común es ((p + 2) (p-2) text <.> ) Multiplicaremos ambos lados de la ecuación por ((p + 2) (p-2) text <:> )

¿La posible solución (p = 2 ) se verifica como una solución real?

La posible solución (p = 2 ) en realidad no funciona, ya que conduce a la división por (0 ) en la ecuación. Entonces, esta es una solución extraña y la ecuación en realidad no tiene solución. Podríamos decir que su conjunto de solución es el conjunto vacío, ( emptyset text <.> )

Ejemplo 12.5.9.

Resuelve (C (t) = 0.35 text <,> ) donde (C (t) = frac <3t>) gives a drug's concentration in milligrams per liter (t) hours since an injection. (This function was explored in the introduction of Section 12.1.)

To solve (C(t)=0.35 ext<,>) we'll begin by setting up (frac<3t>=0.35 ext<.>) We'll begin by identifying that the LCD is (t^2+8 ext<,>) and multiply each side of the equation by this:

This results in a quadratic equation so we will put it in standard form and use the quadratic formula:

Each of these answers should be checked in the original equation they both work. In context, this means that the drug concentration will reach (0.35) milligrams per liter about (1.066) hours after the injection was given, and again (7.506) hours after the injection was given.

Subsection 12.5.2 Solving Rational Equations for a Specific Variable

Rational equations can contain many variables and constants and we can solve for any one of them. The process for solving still involves multiplying each side of the equation by the LCD. Instead of having a numerical answer though, our final result will contain other variables and constants.

Example 12.5.10 .

In physics, when two resistances, (R_1) and (R_2 ext<,>) are connected in a parallel circuit, the combined resistance, (R ext<,>) can be calculated by the formula

Solve for (R) in this formula.

The common denominator is (R R_1 R_2 ext<.>) We will multiply both sides of the equation by (R R_1 R_2 ext<:>)

Example 12.5.11 .

Here is the slope formula

Solve for (x_1) in this formula.

The common denominator is (x_2-x_1 ext<.>) We will multiply both sides of the equation by (x_2-x_1 ext<:>)

Example 12.5.12 .

Solve the rational equation (x=frac<4y-1><2y-3>) for (y ext<.>)

Our first step will be to multiply each side by the LCD, which is simply (2y-3 ext<.>) After that, we'll isolate all terms containing (y ext<,>) factor out (y ext<,>) and then finish solving for that variable.

Subsection 12.5.3 Solving Rational Equations Using Technology

In some instances, it may be difficult to solve rational equations algebraically. We can instead use graphing technology to obtain approximate solutions. Let's look at one such example.

Example 12.5.13 .

Solve the equation (frac<2>=frac<8>) using graphing technology.

We will define (f(x)=frac<2>) and (g(x)=frac<8> ext<,>) and then look for the points of intersection.

Since the two functions intersect at approximately ((-1.524,-0.442)) and ((3.405,4.936) ext<,>) the solutions to (frac<2>=frac<8>) are approximately (-1.524) and (3.405 ext<.>) We can write the solution set as (<-1.524ldots, 3.405ldots>) or in several other forms. It may be important to do something to communicate that these solutions are approximations. Here we used “(ldots)”, but you could also just say in words that the solutions are approximate.

Reading Questions 12.5.4 Reading Questions

Describe what an “extraneous solution” to a rational equation is.

In general, when solving a rational equation, multiplying through by the will leave you with a simpler equation to solve.

When you believe you have the solutions to a rational equation, what is more important than usual (compared to other kinds of equations) for you to do?

Exercises 12.5.5 Exercises

Revisión y calentamiento

Recall the time that Filip traveled with his kids to kick a soccer ball on Mars? We should examine one more angle to our soccer kick question. The formula (H(t)=-6.07t^2+27.1t) finds the height of the soccer ball in feet above the ground at a time (t) seconds after being kicked.

Using technology, find out what the maximum height of the ball was and when it reached that height.

Using technology, solve for when (H(t)=20) and interpret the meaning of this in a complete sentence.

Using technology, solve for when (H(t)=0) and interpret the meaning of this in a complete sentence.

Solving Rational Equations
Solving Rational Equations for a Specific Variable

Solve this equation for (a ext<:>)

Solve this equation for (m ext<:>)

Solve this equation for (x ext<:>)

Solve this equation for (C ext<:>)

Solve this equation for (a ext<:>)

Solve this equation for (c ext<:>)

Solve this equation for (B ext<:>)

Solve this equation for (a ext<:>)

Solving Rational Equations Using Technology

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Application Problems

Scot and Jay are working together to paint a room. If Scot paints the room alone, it would take him (18) hours to complete the job. If Jay paints the room alone, it would take him (12) hours to complete the job. Answer the following question:

If they work together, it would take them hours to complete the job. Use a decimal in your answer if needed.

There are three pipes at a tank. To fill the tank, it would take Pipe A (3) hours, Pipe B (12) hours, and Pipe C (4) hours. Answer the following question:

If all three pipes are turned on, it would take hours to fill the tank.

Casandra and Tien are working together to paint a room. Casandra works (1.5) times as fast as Tien does. If they work together, it took them (9) hours to complete the job. Answer the following questions:

If Casandra paints the room alone, it would take her hours to complete the job.

If Tien paints the room alone, it would take him hours to complete the job.

Two pipes are being used to fill a tank. Pipe A can fill the tank (4.5) times as fast as Pipe B does. When both pipes are turned on, it takes (18) hours to fill the tank. Answer the following questions:

If only Pipe A is turned on, it would take hours to fill the tank.

If only Pipe B is turned on, it would take hours to fill the tank.

Kandace and Jenny worked together to paint a room, and it took them (2) hours to complete the job. If they work alone, it would take Jenny (3) more hours than Kandace to complete the job. Answer the following questions:

If Kandace paints the room alone, it would take her hours to complete the job.

If Jenny paints the room alone, it would take her hours to complete the job.

If both Pipe A and Pipe B are turned on, it would take (2) hours to fill a tank. If each pipe is turned on alone, it takes Pipe B (3) fewer hours than Pipe A to fill the tank. Answer the following questions:

If only Pipe A is turned on, it would take hours to fill the tank.

If only Pipe B is turned on, it would take hours to fill the tank.

Town A and Town B are (570) miles apart. A boat traveled from Town A to Town B, and then back to Town A. Since the river flows from Town B to Town A, the boat’s speed was (25) miles per hour faster when it traveled from Town B to Town A. The whole trip took (19) hours. Answer the following questions:

The boat traveled from Town A to Town B at the speed of miles per hour.

The boat traveled from Town B back to Town A at the speed of miles per hour.

A river flows at (7) miles per hour. A boat traveled with the current from Town A to Town B, which are (260) miles apart. Then, the boat turned around, and traveled against the current to reach Town C, which is (160) miles away from Town B. The second leg of the trip (Town B to Town C) took the same time as the first leg (Town A to Town B). During this whole trip, the boat was driving at a constant still-water speed. Answer the following question:

During this trip, the boat’s speed on still water was miles per hour.

A river flows at (5) miles per hour. A boat traveled with the current from Town A to Town B, which are (100) miles apart. The boat stayed overnight at Town B. The next day, the water’s current stopped, and boat traveled on still water to reach Town C, which is (190) miles away from Town B. The second leg of the trip (Town B to Town C) took (9) hours longer than the first leg (Town A to Town B). During this whole trip, the boat was driving at a constant still-water speed. Find this speed.

Note that you should not consider the unreasonable answer.

During this trip, the boat’s speed on still water was miles per hour.

Town A and Town B are (600) miles apart. With a constant still-water speed, a boat traveled from Town A to Town B, and then back to Town A. During this whole trip, the river flew from Town A to Town B at (20) miles per hour. The whole trip took (16) hours. Answer the following question:

During this trip, the boat’s speed on still water was miles per hour.

Town A and Town B are (350) miles apart. With a constant still-water speed of (24) miles per hour, a boat traveled from Town A to Town B, and then back to Town A. During this whole trip, the river flew from Town B to Town A at a constant speed. The whole trip took (30) hours. Answer the following question:

During this trip, the river’s speed was miles per hour.

Suppose that a large pump can empty a swimming pool in (43 < m hr>) and that a small pump can empty the same pool in (53 < m hr> ext<.>) If both pumps are used at the same time, how long will it take to empty the pool?

If both pumps are used at the same time, it will take to empty the pool.

The winner of a (9 < m mi>) race finishes (14.73 < m min>) ahead of the second-place runner. On average, the winner ran (0.6 < extstylefrac< mmathstrut mi>< mmathstrut hr>>) faster than the second place runner. Find the average running speed for each runner.

The winner's average speed was and the second-place runner's average speed was .

In still water a tugboat can travel (15 < extstylefrac< mmathstrut mi>< mmathstrut hr>> ext<.>) It travels (42 < m mi>) upstream and then (42 < m mi>) downstream in a total time of (5.96 < m hr> ext<.>) Find the speed of the current.

Without any wind an airplane flies at (300 < extstylefrac< mmathstrut mi>< mmathstrut hr>> ext<.>) The plane travels (600 < m mi>) into the wind and then returns with the wind in a total time of (4.04 < m hr> ext<.>) Find the average speed of the wind.

When there is a (11.8 < extstylefrac< mmathstrut mi>< mmathstrut hr>>) wind, an airplane can fly (770 < m mi>) with the wind in the same time that it can fly (702 < m mi>) against the wind. Find the speed of the plane when there is no wind.

It takes one employee (2.5 < m hr>) longer to mow a football field than it does a more experienced employee. Together they can mow the grass in (1.9 < m hr> ext<.>) How long does it take each person to mow the football field working alone?

The more experienced worker takes to mow the field alone, and the less experienced worker takes .

It takes one painter (13 < m hr>) longer to paint a house than it does a more experienced painter. Together they can paint the house in (30 < m hr> ext<.>) How long does it take for each painter to paint the house working alone?

The more experienced painter takes to paint the house alone, and the less experienced painter takes .


Solving Rational Equations Math Maze

Students practice solving algebraic rational equations by completing a math maze!

They draw arrows to show the path that they took to get from the start of the maze to the exit.

This download contains three different mazes for teachers to use to differentiate their instruction.

Level 1 maze questions were designed for students who are struggling to pick up this concept. All problems in this maze can be solved by using the cross product property.

Level 2 maze questions were designed for the average student. In order to solve the problems in this maze, students will first have to multiply by the least common denominator.

Level 3 maze questions were designed for stronger students as an enrichment activity. Prior to multiplying by the least common denominator, students will have to first factor at least one denominator to help them determine what the least common denominator is.


Lane ORCCA (2020–2021): Open Resources for Community College Algebra

Recall from Section 3.3 that we learned the best way to solve equations with fractions is to first clear the fractions by multiplying every term on both sides of the equation by the LCD of all the fractions. Since a rational equation is also an equation with fractions, although with more complicated fractions than those we saw in that earlier section, with variables in the denominators, we can solve rational equations the same way: by first clearing the fractions. You may wish to review the technique of clearing the fractions by eliminating the denominators discussed in Subsection 3.3.2. We will use this same technique to clear the fractions from rational equations.

Example 8.6.1 . Solving a Rational Equation by First Clearing the Fractions.

Use the technique of clearing the fractions to solve the rational equation (frac<2>+frac<1><4>=frac<5><2x> ext<.>)

To clear the fractions from an equation, we must first find the LCD of all the fractions in the equation. Recall that to find the LCD, we should first factor each denominator completely and then take factors from each denominator, one by one, only taking factors that we haven't already taken from a previous denominator, and then multiply them all together.

Here is the equation again with the denominators factored completely:

Now, to build the LCD, we need to take the factor of (x) from the first denominator and both factors of (2) from the second denominator. However, when we look at the third denominator, we notice that we already have both of those factors, so we don't need to take any factors from the third denominator. Thus, the LCD is (xcdot2cdot2=4x ext<.>)

Now, what we do with the LCD is multiply both sides of the equation by the LCD and this will clear the fractions. To show how this works, we'll keep the (4x) and the denominators in factored form. Then, you can see how all of the factors in all of the denominators will end up canceling out.

As you can see, once the fractions are cleared, we end up with a much simpler equation to solve. This is why it is recommended that you clear the fractions first, whenever solving a rational equation.

Remark 8.6.2 .

Because of the step where factors were canceled, it's possible that the solutions obtained through the solving process may not actually be solutions to the original equation. They each might be what is called an . An extraneous solution is a number that would appear to be a solution based on the solving process, but actually does not make the original equation true. This would be the case, for example, if a proposed solution would cause division by zero in the original equation. Because of this, it is important that these proposed solutions always be checked. Note that we're not checking to see if we made a calculation error, but are instead checking to see if the proposed solutions actually solve the original equation or may instead be extraneous.

Let's look at a few more examples of solving rational equations, by first clearing the fractions.

Example 8.6.3 .

The common denominator is (y(y+1) ext<.>) We will multiply both sides of the equation by (y(y+1) ext<:>)

Does the possible solution (y=-3) check as an actual solution?

It checks, so (-3) is a solution. We write the solution set as (<-3> ext<.>)

Example 8.6.4 .

The common denominator is (z-4 ext<.>) We will multiply both sides of the equation by (z-4 ext<:>)

Do the possible solutions (z=1) and (z=4) check as actual solutions?

The possible solution (z=4) does not actually work, since it leads to division by (0) in the original equation. It is an extraneous solution. However, (z=1) is a valid solution. The only solution to the equation is (1 ext<,>) and thus we can write the solution set as (<1> ext<.>)

Example 8.6.5 .

To find the common denominator, we need to factor all denominators first, if possible:

Now we can see the common denominator is ((p+2)(p-2) ext<.>) We will multiply both sides of the equation by ((p+2)(p-2) ext<:>)

Does the possible solution (p=2) check as an actual solution?

The possible solution (p=2) does not actually work, since it leads to division by (0) in original the equation. So this is an extraneous solution and, since there are no other proposed solutions, the equation actually has no solution. We could say that its solution set is the empty set, (emptyset ext<.>)

Subsection 8.6.2 Application Problems

To start this section, we will revisit a scenario we have seen before in Example 8.4.1:

Example 8.6.6 . A Uniform Motion Example.

Julia is taking her family on a boat trip (12) miles down the river and back. The river flows at a speed of (2) miles per hour and she wants to drive the boat at a constant speed, (b) miles per hour downstream and back upstream. Due to the current of the river, the actual speed of travel is (b+2) miles per hour going downstream, and (b-2) miles per hour going upstream. If Julia plans to spend (8) hours for the whole trip, how fast should she drive the boat?

As explained in Example 8.4.1, the time it takes Julia to drive the boat downstream can be represented by (frac<12>) hours, and upstream can be represented by (frac<12>) hours. The function to model the whole trip's time is

where the function name, (t ext<,>) stands for time in hours. The trip will take a total of (8) hours, so we substitute (t(b)) with (8 ext<,>) and we have:

We will solve this equation algebraically, by first clearing the fractions. To remove the fractions in this equation, we will multiply both sides of the equation by the least common denominator ((b-2)(b+2) ext<,>) and we have:

Algebraically, both values do check out to be solutions. However, in the context of this scenario, the boat's speed can't be negative, so we only take the positive solution (4 ext<.>) If Julia drives the boat at (4) miles per hour, the whole trip would take (8) hours.

Let's look at a couple more uniform motion examples.

Example 8.6.7 . Traveling in the Same Amount of Time.

A bus travels (119) miles on a freeway in the same time as it takes a car to travel (153) miles. The bus is traveling (10) mph slower than the car. Find the speed of both the bus and the car.

Let (c) represent the speed of the car, in miles per hour. Then, since the bus is traveling (10) mph slower than the car, the expression (c-10) will represent the speed of the bus, also measured in miles per hour. We can use a table, plus the formula (d=rt ext<,>) to organize our information, similar to Example 5.4.16.

Vehicle
Tipo
Distance Traveled
(in miles)
(=) Rate of Travel
(in mi/hr)
(cdot) Time Traveled
(in hr)
bus (119) (c-10) (t )
carro (153) (c) (t )

Notice that in the problem it says that these distances were traveled by these vehicles in the same time. Thus, if we use the (t=frac) form of the formula for uniform motion ((d=rt)), we can obtain an expression for the time traveled for each vehicle and then set those expressions equal to each other. This will give us a rational equation to solve, which will help us answer the question.

Thus, the car was going (45) mph and the bus, which was going (10) mph slower than the car, was going (35) mph.

Checkpoint 8.6.9 .
Example 8.6.10 . Traveling for a Total Amount of Time.

Jazmine trained for 3 hours on Saturday. She ran (8) miles and then biked (24) miles. Her biking speed is (4) mph faster than her running speed. What is her running speed?

Note that this is also a uniform motion problem, similar to Example 8.6.7. However, it differs in the fact that instead of the times for each part of her workout being the same, we are given the total amount of time. Thus, we can use a similar approach, but instead of setting the expressions representing each time equal to each other, we will add them together and set them equal to the total time.

We will use a table, similar to Table 8.6.8 to organize our information, except we will use expressions to represent each time, since they are not equal and thus shouldn't be represented by the same variable. Let (r) represent Jazmine's running speed. Then, (r+4) must represent her biking speed.

Part of
Workout
Distance Traveled
(in miles)
(=) Rate of Travel
(in mi/hr)
(cdot) Time Traveled
(in hr)
running (8) (r) (frac<8>)
biking (24) (r+4) (frac<24>)

We know that the total time she spent training is (3) hours, so we'll add the expressions representing each time together and set them equal to (3 ext<,>) and finally solve the equation to answer the equation:

Therefore, Jazmine's running speed is (8) mph.

Checkpoint 8.6.12 .

Another type of application problem we will look at is what is known as a . This is a situation where two (or more) people (or machines or pipes, etc.) are completing a task. There are times at which each person can complete the task alone and then there is a time at which they can complete the task if they work together. Any or all of these times may be unknown. Qué es known is the relationship between those times. Here is a formula that describes this relationship, which you may use to set up these shared-work problems:

Here, (a) represents how long it takes one person (or machine or pipe, etc.) to complete a task alone, (b) represents how long it takes a second person (or machine or pipe, etc.) to complete the same task alone, and (c) represents how long it takes for both people (or machines or pipes, etc.) to complete the same task if they work together.

Variables, expressions, or known values may be substituted for (a ext<,>) (b ext<,>) and (c ext<,>) depending on the given situation, and this will give you a rational equation which can be solved to answer the question.

Let's look at a couple shared-work examples.

Example 8.6.13 . A Shared-Work Example.

It takes Ku (3) hours to paint a room and it takes Jacob (6) hours to paint the same room. If they work together, how long would it take them to paint the room?

Since it takes Ku (3) hours to paint the room, we can let (a=3 ext<.>) Similarly, Jacob paints the room in (6) hours so we can let (b=6 ext<.>)

Assume it takes (x) hours to paint the room if Ku and Jacob work together. Then, we will substitute (x) for (c ext<.>) Now we can write this equation:

To clear the fractions, we multiply both sides of the equation by the common denominator of (3 ext<,>) (6) and (x ext<,>) which is (6x ext<:>)

Does the possible solution (x=2) check as an actual solution?

It does, so it is a solution. If Ku and Jacob work together, it would take them (2) hours to paint the room.

Example 8.6.14 .

Pipe A can fill a pool in (3) hours less time than Pipe B can fill a pool. If both pipes are turned on at the same time, then the pool will be filled in (2) hours. How long will it take for each pipe to fill the pool alone?

Let (x) represent the time it takes Pipe B to fill the pool alone. Then, since Pipe A can fill a pool in (3) hours less time than Pipe B can fill a pool, the expression (x-3) will represent the time it takes Pipe A to fill the pool alone. Using the shared-work formula, this gives us the following rational equation:

To solve this equation, we will first clear the fractions by multiplying both sides by the LCD, which is (2x(x-3) ext<,>) and then solve the resulting equation.

This is a quadratic equation, so we will solve this equation by factoring:

Since (x) represents how long it takes Pipe B to fill the pool and Pipe A takes (3) hours less the Pipe B to fill the pool, clearly (1) cannot be a solution. So, it will take Pipe B (6) hours to fill the pool alone and Pipe A (3) hours to fill the pool alone.

Checkpoint 8.6.15 .

Subsection 8.6.3 Solving Rational Equations for a Specific Variable

Rational equations can contain many variables and constants and we can solve for any one of them. The process for solving still involves multiplying each side of the equation by the LCD. Instead of having a numerical answer though, our final result will contain other variables and constants.

Example 8.6.16 .

In physics, when two resistances, (R_1) and (R_2 ext<,>) are connected in a parallel circuit, the combined resistance, (R ext<,>) can be calculated by the formula

Solve for (R) in this formula.

The common denominator is (R R_1 R_2 ext<.>) We will multiply both sides of the equation by (R R_1 R_2 ext<:>)

Example 8.6.17 .

Here is the slope formula

Solve for (x_1) in this formula.

The common denominator is (x_2-x_1 ext<.>) We will multiply both sides of the equation by (x_2-x_1 ext<:>)

Example 8.6.18 .

Solve the rational equation (x=frac<4y-1><2y-3>) for (y ext<.>)

Our first step will be to multiply each side by the LCD, which is simply (2y-3 ext<.>) After that, we'll isolate all terms containing (y ext<,>) factor out (y ext<,>) and then finish solving for that variable.


15.5: Solve Rational Equations - Mathematics

UCI Math 1A/1B: Pre-Calculus
Pre-Calculus: Solving Rational Equations, Part 2
View the complete course: http://ocw.uci.edu/courses/math_1a1b_precalculus.html
Instructor: Sarah Eichhorn, Ph.D and Rachel Lehman, Ph.D

License: Creative Commons CC-BY-SA
Terms of Use: http://ocw.uci.edu/info
More courses at http://ocw.uci.edu

Description: UCI Math 1A/1B: Pre-Calculus is designed to prepare students for a calculus course. This course is taught so that students will acquire a solid foundation in algebra and trigonometry. The course concentrates on the various functions that are important to the study of the calculus.

Required attribution: Eichhorn, Sarah Lehman, Rachel Pre-Calculus 1A/1B (UCI OpenCourseWare: University of California, Irvine), http://ocw.uci.edu/courses/math_1a1b_precalculus.html. [Access date]. License: Creative Commons Attribution-ShareAlike 4.0 United States License.

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Math 1A/1B: Pre-Calculus by Dr. Sarah Eichorn and Dr. Rachel Lehman is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.


Solving Rational Equations Worksheet

1 3 m2 m 4 3m2 2 3m2 2 1 n 1 5n n 1 5n. Show your common denominators and numerators on this sheet or separate paper.

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Solve rational equations by identifying and multiplying by the least common denominator.

Solving rational equations worksheet. Solving Rational Equations Worksheet. 8x 4 X 12 3x q Iox lax 5x x 7 4吘 aox 6x-30 x-5 06. You must check your solutions and throw out any that make the denominator zero.

Solve the resulting equation. Multiply both sides by the LCD. In this resource students will practice solving rational equations.

Create your own worksheets like this one with Infinite Algebra 1. Adding and Subtracting Rational Expressions 2. Recommend to copy the worksheet double-sided since it is 2 pages and then copy the grid separately Once students have solved each problem they will locate the solution in the grid and shade t.

Solving Rational Equations 1 Date_____ Period____ Solve each equation. Only prep work is to make copies. View 38pdf from ALG 1 M31 at Fairview High School Fairview.

Some of the worksheets for this concept are Lesson practice c two step equations with rational numbers Lesson review for mastery two step equations with rational Lesson two step rational inequalities Solving two step equations Solving two step equations with rational coefficients Solve. Solving a Rational Equation 1. 1 a 1 5a 1 a 1 2 6v 6 v2 2 v2 1 v2 3 1 n2 4 n 3 n2 4 4 x 1 x2 1 5ࡨ 5 1 k2 1 3k2 k 5 3k2 6 x 5 x2 1 x 6 x 7 6 k 1 k2 6k 1 k 8 4 n 1.

Practice Worksheet 6 Author. SolveÎ 4 x4 3 x 6. Remember to check for extraneous solutions.

Solving Rational Equations Algebra 2. Solving Rational Equations 2. Create your own worksheets like this one with Infinite Algebra 2.

Find the LCD of all the rational expressions in the equation. About This Quiz Worksheet Solving rational equations is similar to solving equations containing fractions but with an extra stepThis quiz and worksheet combo will test your ability to solve. Remember to check for extraneous solutions.

Solving Rational Equations Solve each equation. 4 x4 3 x 6. LCPS Last modified by.

_ Solving Rational Equations and Complex Fractions Solve each equation. Each one has model problems worked out step by step practice problems challenge proglems and youtube videos that explain each topic. Rational Equations Worksheet With Answers Author.

My students loved this activity as its a fun twist on a difficult concept. Solving Rational Equations Practice Problems Move your mouse over the Answer to reveal the answer or click on the Complete Solution link to reveal all of the steps required for solving rational equations. Students will solve 10 rational equations in which they must use the LCD.

Multiply each side of the equation by 12. Solve Two Step Equations With Rational Numbers – Displaying top 8 worksheets found for this concept. Solving Rational Equations 1.

Solving Rational Equations Free worksheet with answer keys on Rational Expressions-simplifying dividing adding multiplying and more. Solving Rational Equations 1. Remember to check for extraneous solutions.

When solving equations that are made up of rational expressions we will solve them using the same strategy we used to solve linear equations. Solve rational equations with factoring the denominators first Part 2 Duration 445 View the video lesson take notes and complete the problems below. Rational Expression Worksheet Review 16 AddingSubtractingSolving Add or subtract these rational expressions.

Rational Expressions – Solving Rational Equations Objective. Rational Equations Worksheet With Answers Keywords. Adding and Subtracting Rational Expressions 1.

Solve the rational equations. Solving Rational Equations Date_____ Period____ Solve each equation. Notice that we emphasized the must in step 4.

126 4 3 4 12 x x 3x 4 4x 72 3x 12 4x 72 7x 84 x 12 The LCD of the fraction is 12. Rational equations worksheet with answers Created Date. The fractions are eliminated.

1 1 6 k2 1 3k2 1 k 2 1 n2 1 n 1 2n2 3 1 6b2 1 6b 1 b2. Check for extraneous solutions. FACTOR denominators when possible.

Emphasize that each term must be multiplied by the LCD in order to have a balanced equation. Rational equations are equations containing rational expressions. They will then use their answers to solve the math fun fact.

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