Question

Va rog rezolvare completa!
Rezolvati sistemele de ecuatii
a) x=2
y+1=0
b) x+y=5
x=0
c) x+y=3
x+y=4
3) a) 2x-3y+5=0
3x+y+13=0
4) Rezolvati prin metoda substitutiei;a)2(x+y+1)-3(2x-y-3)=12
4(2x-y-1)+3(3-x-y)=3
b) r2× x+r5×y=7
r5×x-r2×y=0
c) x+3\y-1=x\y-8
x-4×y=6
r- este radical
\ -este linie de fractie,supra
Va multumesc

Answer 1

Răspuns:

a) x=2

y=(-1)

b)y=5

x=0

c)∅

3

a) 2x-3y+5=0 2x-3y+5=0 2x-3(-3-13)+5=0

→ →

3x+y+13=0 y= -3x-13 y=-3x-13

x=-4 x=-4

→ →

y= -3(-4)-13 y=-1.

4.

a)

2(x+y+1)-3(2x-y-3)=12 2x+2y+2-6x+3y+9=12...

→

4(2x-y-1)+3(3-x-y)=3 8x-4y-4+9-3x-3y=3.....

.........→ x=1

...... y=1

b)√2× x+√5×y=7→√2× x+√5×y=7→√2×(√10y/5)+√5×y=7→y=√5

√5×x-√2×y=0→ √10y/5 → √10×(√5)/5→x=√2.