CAPITULO XXIV
Ecuaciones simultaneas con dos incognitas
Ecuaciones simultaneas con dos incognitas
- Ejercicio 181
Resolver los siguientes sistemas:
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- { x b + y a =2 x a + y b = a 2 + b 2 ab { ax+by ab =2 bx+ay ab = a 2 + b 2 ab { ax+by=2ab ( 1 ) bx+ay= a 2 + b 2 ( 2 ) Despejo x de ( 1 ) ax+by =2ab ax =2ab–by x = 2ab–by a Reemplazo el valor de x en ( 2 ) b( 2ab–by a ) +ay = a 2 + b 2 b( 2ab–by a ) +ay = a 2 + b 2 2a b 2 – b 2 y a +ay = a 2 + b 2 2 a b 2 a – b 2 y a +ay = a 2 + b 2 y( a– b 2 a ) = a 2 –2 b 2 + b 2 y( a 2 – b 2 a ) = a 2 – b 2 y ( a 2 – b 2 ) =a ( a 2 – b 2 ) y =a Reemplazo el valor de y en ( 1 ) ax+by =2ab ax+ab =2ab ax =2ab–ab a x = a b x =b Sol.{ x=b y=a
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- { ( a–b ) x–( a+b ) y= b 2 –3ab ( a+b ) x–( a–b ) y=ab– b 2 { ( a–b ) x= b 2 –3ab+( a+b ) y ( a+b ) x=ab– b 2 +( a–b ) y { x= b 2 –3ab+( a+b ) y a–b ( 1 ) x= ab– b 2 +( a–b ) y a+b ( 2 ) ( 1 ) =( 2 ) b 2 –3ab+( a+b ) y a–b = ab– b 2 +( a–b ) y a+b b 2 –3ab a–b + ( a+b ) y a–b = ab– b 2 a+b + ( a–b ) y a+b ( a+b ) y a–b – ( a–b ) y a+b = ab– b 2 a+b – b 2 –3ab a–b y( a+b a–b – a–b a+b ) = b( a–b ) a+b – b( b–3a ) a–b y[ ( a+b ) 2 – ( a–b ) 2 ( a–b ) ( a+b ) ] =b[ a–b a+b – b–3a a–b ] y[ a 2 +2ab+ b 2 – a 2 +2ab– b 2 ( a–b ) ( a+b ) ] =b[ ( a–b ) 2 –( a+b ) ( b–3a ) ( a–b ) ( a+b ) ] y( 4a b ) = b [ a 2 –2ab+ b 2 –( ab–3 a 2 + b 2 –3ab ) ] 4ay = a 2 – 2ab + b 2 +3 a 2 – b 2 + 2ab 4ay =4 a 2 y = 4 a 2 4 a y =a Reemplazo el valor de y en ( 2 ) x = ab– b 2 +( a–b ) y a+b x = ab– b 2 +( a–b ) a a+b x = ab – b 2 + a 2 – ab a+b x = a 2 – b 2 a+b x = ( a+b ) ( a–b ) a+b x =a–b Sol.{ x=a–b y=a
- { x+b a + y–b b = a+b b x–a b – y–a a =– a+b a { x+b a = a+b b – y–b b x–a b = y–a a – a+b a { x=a( a+b b – y–b b ) –b ( 1 ) x=b( y–a a – a+b a ) +a ( 2 ) ( 1 ) =( 2 ) a( a+b b – y–b b ) –b =b( y–a a – a+b a ) +a a b ( a+b–y+b ) –b = b a ( y–a–a–b ) +a a b ( a+2b–y ) –b = b a ( y–2a–b ) +a a 2 b +2a– ay b –b = by a –2b– b 2 a +a a 2 b +2a–b+2b+ b 2 a –a = ay b + by a a 2 b +a+b+ b 2 a =y( a b + b a ) a 3 + a 2 b+a b 2 + b 3 ab =y( a 2 + b 2 ab ) a 2 ( a+b ) + b 2 ( a+b ) =y( a 2 + b 2 ) ( a 2 + b 2 ) ( a+b ) =y ( a 2 + b 2 ) y =a+b Reemplazo el valor de y en ( 1 ) x =a( a+b b – y–b b ) –b x =a( a+b b – a+ b – b b ) –b x = a b ( a +b– a ) –b x = a b ( b ) –b x =a–b Sol.{ x=a–b y=a+b
- { x a+b + y a+b = 1 ab x b + y a = a 2 + b 2 a 2 b 2 { x a+b = 1 ab – y a+b x b = a 2 + b 2 a 2 b 2 – y a { x= a+b ab –y ( 1 ) x= a 2 + b 2 a 2 b – by a ( 2 ) ( 1 ) =( 2 ) a+b ab –y = a 2 + b 2 a 2 b – by a by a –y = a 2 + b 2 a 2 b – a+b ab y( b a –1 ) = 1 ab [ a 2 + b 2 a –( a+b ) ] y( b–a a ) = 1 a b [ a 2 + b 2 –a( a+b ) a ] y( b–a ) = 1 b [ a 2 + b 2 – a 2 –ab a ] y( b–a ) = 1 b × b ( b–a ) a y = ( b–a ) a ( b–a ) y = 1 a Reemplazo el valor de y en ( 1 ) x = a+b ab –y x = a+b ab – 1 a x = 1 a ( a+b b –1 ) x = 1 a ( a+ b – b b ) x = 1 a ( a b ) x = 1 b Sol.{ x= 1 b y= 1 a
