Icono del sitio Solucionario Baldor

Ejercicio 177

Gestor Baldor
Comparte esto 👍👍
DESCARGA
CAPITULO XXIV

Ecuaciones simultaneas con dos incognitas
Ejercicio 177
Resolver por el método de igualación:
  1. { x+3y=6 ( 1 ) 5x–2y=13 ( 2 ) Despejando x de ( 1 ) x =6–3y Reemplazo x en ( 2 ) 5x–2y =13 5( 6–3y ) –2y =13 30–15y–2y =13 –17y =–30+13 – 17 y =– 17 y =1 Reemplazo y en ( 1 ) x =6–3y x =6–3( 1 ) x =3 Sol.{ x=3 y=1
  2. { 5x+7y=–1 ( 1 ) –3x+4y=–24 ( 2 ) Despejando x de ( 1 ) 5x =–7y–1 x = –7y–1 5 x =– 7y+1 5 Reemplazo el valor de x en ( 2 ) –3x+4y =–24 –3( – 7y+1 5 ) +4y =–24 3( 7y+1 ) 5 =–4y–24 21y+3 =5( –4y–24 ) 21y+3 =–20y–120 21y+20y =–3–120 41y =–123 y = – 41 y =–3 Reemplazo y en ( 2 ) –3x+4y =–24 –3x+4( –3 ) =–24 –3x–12 =–24 –3x =12–24 –3x =–12 x = – – 3 x =4 Sol.{ x=4 y=–3
  3. { 4y+3x=8 ( 1 ) 8x–9y=–77 ( 2 ) Despejando x de ( 1 ) 3x =8–4y 3x =4( 2–y ) x = 4( 2–y ) 3 Reemplazando el valor de x en ( 2 ) 8x–9y =–77 8 4( 2–y ) 3 –9y =–77 32( 2–y ) 3 =9y–77 32( 2–y ) =3( 9y–77 ) 64–32y =27y–231 –27y–32y =–64–231 –59y =–295 y = – – 59 y =5 Reemplazo y en ( 1 ) 4y+3x =8 4( 5 ) +3x =8 20+3x =8 3x =8–20 3x =–12 x =– 3 x =–4 Sol.{ x=–4 y=5
  4. { x–5y=8 ( 1 ) –7x+8y=25 ( 2 ) Despejando x de ( 1 ) x–5y =8 x =8+5y Reemplazo el valor de x en ( 2 ) –7x+8y =25 –7( 8+5y ) +8y =25 –56–35y+8y =25 –27y =56+25 –27y =81 y =– 27 y =–3 Reemplazo y en ( 1 ) x–5y =8 x–5( –3 ) =8 x+15 =8 x =8–15 x =–7 Sol.{ x=–7 y=–3
  5. { 15x+11y=32 ( 1 ) 7y–9x=8 ( 2 ) Despejando y de ( 1 ) 15x+11y =32 11y =32–15x y = 32–15x 11 Reemplazando el valor de y en ( 2 ) 7y–9x =8 7( 32–15x 11 ) –9x =8 7( 32–15x 11 ) =9x+8 7( 32–15x ) =11( 9x+8 ) 224–105x =99x+88 –105x–99x =–224+88 –204x =–136 x = – – x = 2 3 Reemplazo x en ( 2 ) 7y–9x =8 7y–( 2 3 ) =8 7y–6 =8 7y =6+8 7y =14 y = 7 y =2 Sol.{ x= 2 3 y=2
  6. { 10x+18y=–11 ( 1 ) 16x–9y=–5 ( 2 ) Despejando x de ( 1 ) 10x+18y =–11 10x =–18y–11 x =– 18y+11 10 Reemplazando el valor de x en ( 2 ) 16x–9y =–5 ( – 18y+11 ) –9y =–5 8( – 18y+11 5 ) =9y–5 –8( 18y+11 ) =5( 9y–5 ) –144y–88 =45y–25 –144y–45y =88–25 –189y =63 y = 63 – y =– 1 3 Reemplazo y en ( 2 ) 16x–9y =–5 16x–( – 1 3 ) =–5 16x+3 =–5 16x =–5–3 x = – 8 x =– 1 2 Sol.{ x=– 1 2 y=– 1 3
  7. { 4x+5y=5 ( 1 ) –10y–4x=–7 ( 2 ) Despejando x de ( 1 ) 4x+5y =5 4x =5–5y 4x =5( 1–y ) x = 5( 1–y ) 4 Reemplazando el valor de x en ( 2 ) –10y–4x =–7 –10y– 4 [ 5( 1–y ) 4 ] =–7 –10y–5( 1–y ) =–7 –10y–5+5y =–7 –5y =5–7 –5y =–2 y = 2 5 Reemplazo y en ( 1 ) 4x+5y =5 4x+ 5 ( 2 5 ) =5 4x+2 =5 4x =5–2 4x =3 x = 3 4 Sol.{ x= 3 4 y= 2 5
  8. { 32x–25y=13 ( 1 ) 16x+15y=1 ( 2 ) Despejando x de ( 2 ) 16x+15y =1 16x =1–15y x = 1–15y 16 Reemplazando el valor de x en ( 1 ) 32x–25y =13 ( 1–15y 16 ) –25y =13 2( 1–15y ) =25y+13 2–30y =25y+13 –25y–30y =13–2 –55y =11 y =– 11 y =– 1 5 Reemplazo y en ( 2 ) 16x+15y =1 16x+( – 1 5 ) =1 16x–3 =1 16x =4 x = 4 x = 1 4 Sol.{ x= 1 4 y=– 1 5
  9. { –13y+11x=–163 ( 1 ) –8x+7y=94 ( 2 ) Despejando x de ( 2 ) –8x+7y =94 –8x =94–7y x = 7y–94 8 Reemplazando el valor x en ( 1 ) –13y+11x =–163 –13y+11( 7y–94 8 ) =–163 11( 7y–94 8 ) =13y–163 11( 7y–94 ) =8( 13y–163 ) 77y–1034 =104y–1304 77y–104y =1034–1304 –27y =–270 y = – – 27 y =10 Reemplazo y en ( 2 ) –8x+7y =94 –8x+7( 10 ) =94 –8x+70 =94 –8x =94–70 –8x =24 x =– 8 x =–3 Sol.{ x=–3 y=10
Salir de la versión móvil