Question

Rezolvaţi prin metoda substituţiei: 2(x+y+1)-3(2x - y-3)=12 14(2x-y-1)+3(3-x - y)=3 x + 3 X b) | $2.x+√5.y=7 1/5.x-√2.y=0 c) y-1 -8 1-4.y = 6




Va rog repede e urgent!!!Dau 15puncte și coroana!!!​

Answer 1

Răspuns:

Explicație pas cu pas:

2(x+y+1)-3(2x-y-3)=12 ⇒2x+2y+2-6x+3y+9=12  ⇒-4x+5y=12-11 ⇒-4x+5y=1

4(2x-y-1)+3(3-x - y)=3  ⇒8x-4y-4+9-3x-3y=3 ⇒5x-7y=-2  

-4x+5y=1  ⇒5y=1+4x  ⇒y=(1+4x)/5  

5x-7y=-2         ⇒5x-7·(1+4x)/5 =-2  ⇒25x-7-28x=-10  ⇒-3x=-10+7  ⇒-3x=-3  ⇒x=1    ⇒y=(1+4x)/5  ⇒y=(1+4)/5=1 y=1

b) √2·x+√5·y=7                                          √2·√2y/√5 +√5·y=7  ⇒

    √5·x-√2·y=0 ⇒√5·x=√2·y  ⇒x=√2y/√5   Inlocuim in prima relatie ⇒√2·√2y/√5 +√5·y=7  ⇒2√5y/5+5√5y/5=7  ⇒y(2√5+5√5)=35  ⇒y·7√5=35 I:7  ⇒y·√5=5  ⇒y=5/√5 =5√5/5=√5  Inlocuim in relati pt. x

x=√2y/√5=√2·√5/√5=√2    x=√2   y=√5

c) (x+3)/(y-1)= x/(y-8)  ⇒(x+3)·(y-8)=x·(y-1)  ⇒

    x-4·y = 6  ⇒x=4y+6    Inlocuim in prima relatie.

(x+3)·(y-8)=x·(y-1)  ⇒(4y+6 +3)·(y-8)=(4y+6)·(y-1)  ⇒(4y+9)·(y-8)=(4y+6)·(y-1)  ⇒

4y²-32y+6y-6=4y²+6y-4y+8  ⇒4y²-26y-4y²-6y=8+6  ⇒-32y=14  ⇒y=-14/32=-7/16   y=-7/16  

x=4y+6=4·(-7/16)+6= -7/4+24/4=17/4  x=17/4