Rozwiąż układy równań:
a){(x+3)^2-(y-`)^2=(x-y)(x+y)
3x-7y=0
b) {2x-y/4 +3x+5/2=3
x-7/5-2y+x/10=-1
w a powinno wyjść x= - 1 i 1/6 y=-1/2
w b powinno wyjsc x=0 y=-2
Z góry bardzo dziękuję =D
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
(x+3)2-(y-1)2=(x-y)(x+y)
3x-7y=0
x2+6x+9-y2+2y-1=x2-y2
3x-7y=0
x2-x2-y2+y2+6x+2y=-8
3x-7y=0
6x+2y=-8
3x-7y=0 / * (-2)
6x+2y=-8
-6x+14y=0
.................
16y=-8
y=- 1/2
6x+2 * (-1/2) = -8
6x= -8+1
x=- 7/6=- 1 1/6
x= - 1 1/6
y= - 1/2
b) {2x-y/4 +3x+5/2=3 / * 4
x-7/5-2y+x/10=-1 /* 10
2x-y+2(3x+5)=12
2(x-7)-2y-x=-10
2x-y+6x+10=12
2x-14-2y-x=-10
-y=-8x+2
x-2y=4
y=8x-2
x-2(8x-2)=4
y=8x-2
x-16x+4=4
y=8x-2
-15x=0
y=8x-2
x=0
y=8 * 0 - 2 = - 2
x=0
x=0
y= -2