Ejercicio 181

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CAPITULO XXIV

Ecuaciones simultaneas con dos incognitas
Ejercicio 181
Resolver los siguientes sistemas:
  1. Mathematical Equation
  2. Mathematical Equation
  3. Mathematical Equation
  4. Mathematical Equation
  5. Mathematical Equation
  6. { 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 =2abby x = 2abby a Reemplazo el valor de x en ( 2 ) b( 2abby a ) +ay = a 2 + b 2 b( 2abby 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 =2abab a x = a b x =b Sol.{ x=b y=a
  7. Mathematical Equation
  8. Mathematical Equation
  9. Mathematical Equation
  10. Mathematical Equation
  11. Mathematical Equation
  12. Mathematical Equation
  13. Mathematical Equation
  14. Mathematical Equation
  15. Mathematical Equation
  16. Mathematical Equation
  17. Mathematical Equation
  18. { (ab ) x(a+b ) y= b 2 3ab (a+b ) x(ab ) y=ab b 2 { (ab ) x= b 2 3ab+(a+b ) y (a+b ) x=ab b 2 +(ab ) y { x= b 2 3ab+(a+b ) y ab ( 1 ) x= ab b 2 +(ab ) y a+b ( 2 ) ( 1 ) =( 2 ) b 2 3ab+(a+b ) y ab = ab b 2 +(ab ) y a+b b 2 3ab ab + (a+b ) y ab = ab b 2 a+b + (ab ) y a+b (a+b ) y ab (ab ) y a+b = ab b 2 a+b b 2 3ab ab y( a+b ab ab a+b ) = b(ab ) a+b b(b3a ) ab y[ (a+b ) 2 (ab ) 2 (ab ) (a+b ) ] =b[ ab a+b b3a ab ] y[ a 2 +2ab+ b 2 a 2 +2ab b 2 (ab ) (a+b ) ] =b[ (ab ) 2 (a+b ) (b3a ) (ab ) (a+b ) ] y(4a b ) = b [ a 2 2ab+ b 2 (ab3 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 +(ab ) y a+b x = ab b 2 +(ab ) a a+b x = ab b 2 + a 2 ab a+b x = a 2 b 2 a+b x = (a+b )(ab ) a+b x =ab Sol.{ x=ab y=a
  19. { x+b a + yb b = a+b b xa b ya a = a+b a { x+b a = a+b b yb b xa b = ya a a+b a { x=a( a+b b yb b ) b ( 1 ) x=b( ya a a+b a ) +a ( 2 ) ( 1 ) =( 2 ) a( a+b b yb b ) b =b( ya a a+b a ) +a a b (a+by+b ) b = b a (yaab ) +a a b (a+2by ) b = b a (y2ab ) +a a 2 b +2a ay b b = by a 2b b 2 a +a a 2 b +2ab+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 yb 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 =ab Sol.{ x=ab y=a+b
  20. { 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( ba a ) = 1 a b [ a 2 + b 2 a(a+b ) a ] y(ba ) = 1 b [ a 2 + b 2 a 2 ab a ] y(ba ) = 1 b × b (ba ) a y = (ba ) a (ba ) 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