Question

Rozwiąż układy równan:

a) (x+1)²-(y-2)=x²+2(y-1)
3(x-2)+y²=(y+1)(y-1)

b) 2x-1/5 +3y+2/4 =2
3x+1/5-3y+2/4=0

/ - kreska ulamkowa

Answer 1

A)
(x+1)²-(y-2)=x²+2(y-1)
3(x-2)+y²=(y+1)(y-1)

x² + 2x + 1 - y + 2 = x²+2y - 2
3x - 6 +y² = y² - 1

x² - x² -2y + 2x - y = - 2 - 2 - 1
3x + y² - y² = -1 + 6

3y + 2x = - 5
3x = 5 / : 3

3y + 2x = - 5
x = 5/3


3y + 2 * 5/3 = -5
3y = -5 - 7/3
3y = - 7 całych i 1/3 / :3
y = - 2 całe i 4/9


b)
2x-1/5 +3y+2/4 =2
3x+1/5-3y+2/4=0

2x + 3y = 2 + 1/5 - 1/2
3x - 3y = - 1/5 - 1/2

2x + 3y = 1,7
3x - 3y = - 0,7

5x = 1 / : 5
x= 1/5

2x + 3y = 1,7
2 * 1/5 + 3y = 1,7
3y = 1,7 - 0,4
3y = 1,3 /:3
y = 13/30