MATH 1050 | UNIDAD 6 | 6.1 EJERCICIOS DE APLICACIN Nombre: ________________________________________ ______________________ Fecha: Instrucciones: Estudia y repasa el material de estudio antes de realizar este trabajo. En este, vas a resolver ejercicios sobre ecuaciones, sistemas de ecuaciones y desigualdades lineales en dos variables. Debes presentar el procedimiento de cada ejercicio para poder obtener puntuacin completa en cada uno. Recuerda enviar usando el formato APA. I. II. Resuelve los siguientes sistemas de ecuaciones lineales utilizando el mtodo de sustitucin. (5 pts. c/u) 1. y=10 {xx+2 2 y =6 2. x +3 y=15 {2x+2 y=4 3. =4 {62x x++3 yy=12 Resuelve los siguientes sistemas de ecuaciones lineales utilizando el mtodo de eliminacin. (5 pts. c/u) 1. x+2 y =6 {53x4 y=10 2. {105 xx+36y=6 y=1 2 MATH 1050 3. III. IV. | UNIDAD 6 | 6.1 EJERCICIOS DE APLICACIN {26 x+x+73 y=15 y=21 Resuelve los siguientes sistemas de desigualdades lineales utilizando el mtodo grfico. (5 pts. c/u) 1. {32x+x +4yy<3<8 2. y 14 {3 x7 x +5 y <10 determina el punto de equilibrio en los siguientes problemas. explica, con al menos 2 oraciones completas, qu significado tiene este cada una las situaciones. (5 pts. c u) 1. costosfijos por producir cierto artculo son $5,495 mes y variables $4.50 unidad. si productor vende uno artculos a $8.00, explica equilibro. 2. costo x est dado donde yc es total producido se $7.00. a. encuentra equilibrio. (2 pts) x+ 1050 artculos. , math | unidad 6 6.1 ejercicios aplicacin b. sabe que 175 unidades vendern, cul debera ser precio fijado para garantizar no haya prdidas? (3 pts.) 3. la semana . puede venderse fabricante reducir $4 incrementando $1,640 semana, le convendr hacerlo? i 1) x+2y =10 x-2y =-6 solution: from first equation ... (i) on putting the value into second equation, we get 10-2y-2y =-6 4y =16 in (i),>y=0 From (i); 3x+2(0) =6 =>x=2 Therefore, x=2, y=0 ................................ 2) Solution: 5x-3y=6 ... (i) -10x+6y= 12 ... (ii) Multiply (i) by 2 and (ii) by 1, we get 10x-6y=12 -10x+6y=12 Adding both equations, we get 0=24, it is impossible. =>y=0 Hence, no solution exists. ..................................... 3) Solution: 6x+3y=15 ... (i) -2x+7y= -21 ... (ii) Multiply (i) by 1 and (ii) by 3, we get 6x+3y=15 ... (i) -6x+21y= -63 ... (ii) Adding both equations, we get 24y=-48 => y=-2 From (i); 6x+3(-2)=15 => 6x=21 =>x=7/2 Therefore, x=7/2, y=-2 ................................ III 1) 2x+y<3 -3x+4y<-8 .......................................................... 2) 3x-7y -14 x+5y<10 solution: ........................................................... iv 1) let c(x) and r(x) be cost revenue function respectively. where x number of items. then, for break-even point 5495+4.5x=8x 3.5x =5495 it means 1570 items are required to cover total costs, consisting both fixed variable costs the company. ................................................. here is yc 2.8 1050 function. then 2.8x+1050 =7x 4.2x =1050 250 if 175 sold, manufacturing them 1540 no losses price item should set .................................................... 3) , c =1500+5x 1500+5x =8x 3x =1500 500 new 1640+4x=8x 4x=1640 yes, because just at 410 as compared ................................. i x+2y =10 x-2y =-6 from first equation ... (i) on putting value into second equation, we get 10-2y-2y =-6 4y =16 y =4 in (i),>y=0 From (i); 3x+2(0) =6 =>x=2 Therefore, x=2, y=0 ................................ 2) Solution: 5x-3y=6 ... (i) -10x+6y= 12 ... (ii) Multiply (i) by 2 and (ii) by 1, we get 10x-6y=12 -10x+6y=12 Adding both equations, we get 0=24, it is impossible. =>y=0 Hence, no solution exists. ..................................... 3) Solution: 6x+3y=15 ... (i) -2x+7y= -21 ... (ii) Multiply (i) by 1 and (ii) by 3, we get 6x+3y=15 ... (i) -6x+21y= -63 ... (ii) Adding both equations, we get 24y=-48 => y=-2 From (i); 6x+3(-2)=15 => 6x=21 =>x=7/2 Therefore, x=7/2, y=-2 ................................ III 1) 2x+y<3 -3x+4y<-8 .......................................................... 2) 3x-7y -14 x+5y<10 solution: ........................................................... iv 1) let c(x) and r(x) be cost revenue function respectively. where x number of items. then, for break-even point 5495+4.5x=8x 3.5x =5495 it means 1570 items are required to cover total costs, consisting both fixed variable costs the company. ................................................. here is yc 2.8x 1050 function. then 2.8x+1050 =7x 4.2x =1050 250 if 175 sold, manufacturing them 2.8 1540 no losses price item should set .................................................... 3) , c =1500+5x 1500+5x =8x 3x=1500 500 new 1640+4x =8x 4x =1640 yes, because just at 410 as compared ................................. i x+2y =10 x-2y =-6 from first equation ... (i) on putting value into second equation, we get 10-2y-2y =-6 4y =16 y =4 in (i),>y=0 From (i); 3x+2(0) =6 =>x=2 Therefore, x=2, y=0 ................................ 2) Solution: 5x-3y=6 ... (i) -10x+6y= 12 ... (ii) Multiply (i) by 2 and (ii) by 1, we get 10x-6y=12 -10x+6y=12 Adding both equations, we get 0=24, it is impossible. =>y=0 Hence, no solution exists. ..................................... 3) Solution: 6x+3y=15 ... (i) -2x+7y= -21 ... (ii) Multiply (i) by 1 and (ii) by 3, we get 6x+3y=15 ... (i) -6x+21y= -63 ... (ii) Adding both equations, we get 24y=-48 => y=-2 From (i); 6x+3(-2)=15 => 6x=21 =>x=7/2 Therefore, x=7/2, y=-2 ................................ III 1) 2x+y<3 -3x+4y<-8 .......................................................... 2) 3x-7y -14 x+5y<10 solution: ........................................................... iv 1) let c(x) and r(x) be cost revenue function respectively. where x number of items. then, for break-even point 5495+4.5x=8x 3.5x =5495 it means 1570 items are required to cover total costs, consisting both fixed variable costs the company. ................................................. here is yc 2.8 1050 function. then 2.8x+1050 =7x 4.2x =1050 250 if 175 sold, manufacturing them 1540 no losses price item should set .................................................... 3) , c =1500+5x 1500+5x =8x 3x =1500 500 new 1640+4x=8x 4x=1640 yes, because just at 410 as compared ................................. i x+2y =10 x-2y =-6 from first equation ... (i) on putting value into second equation, we get 10-2y-2y =-6 4y =16 y =4 in (i),>y=0 From (i); 3x+2(0) =6 =>x=2 Therefore, x=2, y=0 ................................ 2) Solution: 5x-3y=6 ... (i) -10x+6y= 12 ... (ii) Multiply (i) by 2 and (ii) by 1, we get 10x-6y=12 -10x+6y=12 Adding both equations, we get 0=24, it is impossible. =>y=0 Hence, no solution exists. ..................................... 3) Solution: 6x+3y=15 ... (i) -2x+7y= -21 ... (ii) Multiply (i) by 1 and (ii) by 3, we get 6x+3y=15 ... (i) -6x+21y= -63 ... (ii) Adding both equations, we get 24y=-48 => y=-2 From (i); 6x+3(-2)=15 => 6x=21 =>x=7/2 Therefore, x=7/2, y=-2 ................................ III 1) 2x+y<3 items